It’s easier than it sounds; let’s start with an example (click here):

Type in the logarithmic equation

Press Go

Another example (click here):

Stay tuned for more.

Cheers,

Michal

Logarithmic equations are often used in various scales of measurement; reducing wide-ranging quantities to smaller scopes. Examples of logarithmic equations applications in real life are with earthquakes scale (Richter), sound scale (decibel), pH levels, radioactive decay, bacterial growth, population growth, continuous interest and more. Logarithmic equations are pretty useful; you might as well learn how to solve them.

To solve logarithmic equations you should first manipulate the equation to have log with the same base on both sides so you can drop the logs (the only way two logs can be equal if their arguments are equal). The next step is solving a linear or polynomial equation (we know how to do that), last but not least you should check the solutions by plugging them into the original equation (we can only plug positive numbers into a logarithm).

It’s easier than it sounds; let’s start with an example (click here):

Type in the logarithmic equation

Press Go

Another example (click here):

Stay tuned for more.

Cheers,

Michal

It’s easier than it sounds; let’s start with an example (click here):

Type in the logarithmic equation

Press Go

Another example (click here):

Stay tuned for more.

Cheers,

Michal

Parallel lines have the same slope, to find the parallel line at a given point you should simply calculate the slope, calculate the y intercept, and you’re pretty much done.

Finding the equation of the perpendicular line is somewhat similar, only calculating the slope is trickier. The slope of the perpendicular line is the negative reciprocal of the line slope, in other words, when multiplying the slopes of two perpendicular lines you should get -1.

Here’s how to find the equation of the line that is parallel to a given line and passes through a given point (click here):

In short, type in Parallel, the line and the point:

Click Go:

Here’s how to find the equation of the line that is perpendicular to a given line and passes through a given point (click here):

Type in Perpendicular, the line and the point:

Click Go:

Cheers,

Michal

Finding the equation of the perpendicular line is somewhat similar, only calculating the slope is trickier. The slope of the perpendicular line is the negative reciprocal of the line slope, in other words, when multiplying the slopes of two perpendicular lines you should get -1.

Here’s how to find the equation of the line that is parallel to a given line and passes through a given point (click here):

In short, type in Parallel, the line and the point:

Click Go:

Here’s how to find the equation of the line that is perpendicular to a given line and passes through a given point (click here):

Type in Perpendicular, the line and the point:

Click Go:

Cheers,

Michal

A biquadratic equation is a quadratic function of a square, having the form f(x)= ax^4+ bx^2 + c. To solve you simply have to rewrite the equation as a quadratic equation: substitute x^4 with u^2 and x^2 with u, to get au^2+ bu + c. See previous post on solving quadratic functions here.

Don’t forget to substitute back and solve for x. You should get up to 4 solutions; it is a quartic equation after all.

Let’s see how it works (click here):

Here’s another example (click here):

Cheers,

Michal

Don’t forget to substitute back and solve for x. You should get up to 4 solutions; it is a quartic equation after all.

Let’s see how it works (click here):

Here’s another example (click here):

Cheers,

Michal

Radical equations are equations involving radicals of any order. We will show examples of square roots; higher order radicals simply require more of the same work. To solve radical equations you first have to get rid of the radicals, in the case of square roots square both sides of the equation (in some cases this should be done multiple times), then refine the new equation (either linear or quadratic) and solve. One more thing to note, by squaring the equation we changed the original equation, so it is very important to check the solutions at the end.

Let’s see how it works, simply follow the steps:

1. Simplify: eliminate the square root

2. Refine: refine the quadratic equation after squaring the root

3. Solve: solve the equation

4. Verify: plug the solutions into the original equation and verify

Click here for an example:

Now let’s zoom in on how to simplify the radical. You will notice that the first step is to isolate the square root before squaring both sides of the equation:

Here’s a more advanced example that requires multiple steps to resolve the radical (click here):

Cheers,

Michal

Let’s see how it works, simply follow the steps:

1. Simplify: eliminate the square root

2. Refine: refine the quadratic equation after squaring the root

3. Solve: solve the equation

4. Verify: plug the solutions into the original equation and verify

Click here for an example:

Now let’s zoom in on how to simplify the radical. You will notice that the first step is to isolate the square root before squaring both sides of the equation:

Here’s a more advanced example that requires multiple steps to resolve the radical (click here):

Cheers,

Michal

What’s in common between Perpendicular Line, Normal Line, Elimination Method and Bi-quadratic Equations? All are new features we’ve just added. This calls for a blog post!

So what’s new? Step by step solutions for Normal line, Perpendicular line, Elimination Method, and Biquadratic Equations, pretty amazing!

Let’s dive right into this:

**Perpendicular Line**

Perpendicular lines have a negative reciprocal slope, all it takes to find the equation of the line perpendicular to the line at a point is to calculate the slope.

Let’s see how it works (click here):

**Normal Line**

The normal line to a curve at a given point is the line perpendicular to the tangent line to the curve at the point. To find the normal line, find the tangent line first then find the equation of the perpendicular line (same technique as above). Symbolab knows how to do all this, type Normal (along with the function and point), and Go.

Here’s how it looks like (click here):

**System of Equations – Elimination Method**

The Elimination Method is the process of eliminating one of the variables in a system of equations using addition or subtraction. The Substitution Method is the process of solving an equation for one variable, and subsequently substituting that solution in the other equation. In most cases you can use one or the other.

We’ve made it easier for you to select the method of your choice. Type in the system of equations, press Go, you should notice the different methods listed.

Click here to see how it works:

Simply click “Using the elimination method” to get the solution steps by elimination:

**Bi-quadratic Equations**

A bi-quadratic equation is a quadratic function of a square, having the form f(x)=ax^4+bx^2+c. To solve you simply have to rewrite the equation as a quadratic equation.

Here’s how Symbolab does just that (click here):

Enjoy the new features!

Cheers,

Michal

So what’s new? Step by step solutions for Normal line, Perpendicular line, Elimination Method, and Biquadratic Equations, pretty amazing!

Let’s dive right into this:

Let’s see how it works (click here):

The normal line to a curve at a given point is the line perpendicular to the tangent line to the curve at the point. To find the normal line, find the tangent line first then find the equation of the perpendicular line (same technique as above). Symbolab knows how to do all this, type Normal (along with the function and point), and Go.

Here’s how it looks like (click here):

The Elimination Method is the process of eliminating one of the variables in a system of equations using addition or subtraction. The Substitution Method is the process of solving an equation for one variable, and subsequently substituting that solution in the other equation. In most cases you can use one or the other.

We’ve made it easier for you to select the method of your choice. Type in the system of equations, press Go, you should notice the different methods listed.

Click here to see how it works:

Simply click “Using the elimination method” to get the solution steps by elimination:

A bi-quadratic equation is a quadratic function of a square, having the form f(x)=ax^4+bx^2+c. To solve you simply have to rewrite the equation as a quadratic equation.

Here’s how Symbolab does just that (click here):

Enjoy the new features!

Cheers,

Michal

On the last post we covered completing the square (see link). It is pretty strait forward if you follow all the steps (there are quite a few steps).

To make things simple, a general formula can be derived such that for a quadratic equation of the form ax²+bx+c=0 the solutions are x=(-b ± sqrt(b^2-4ac))/2a. The quadratic formula comes in handy, all you need to do is to plug in the coefficients and the constants (a,b and c). One thing to note, you must memorize the formula, it is not as intuitive as factoring or completing the square.

Let’s see how it works (click here):

Here’s an advanced example that involves complex numbers (click here):

We’ve got you covered; that was the last on the Quadratic Equations series. Now you know everything there is to know about solving quadratic equations.

Cheers,

Michal

To make things simple, a general formula can be derived such that for a quadratic equation of the form ax²+bx+c=0 the solutions are x=(-b ± sqrt(b^2-4ac))/2a. The quadratic formula comes in handy, all you need to do is to plug in the coefficients and the constants (a,b and c). One thing to note, you must memorize the formula, it is not as intuitive as factoring or completing the square.

Let’s see how it works (click here):

Here’s an advanced example that involves complex numbers (click here):

We’ve got you covered; that was the last on the Quadratic Equations series. Now you know everything there is to know about solving quadratic equations.

Cheers,

Michal

Solving quadratics by factorizing (link to previous post) usually works just fine. But what if the quadratic equation can’t be factored, you're going to need a different strategy to help you solve it.

An equation in which one side is a perfect square trinomial can be easily solved by taking the square root of each side. Easy is good, so we basically want to force the quadratic equation into the form (x+a)²=x²+2ax+a².

All it takes is making sure that the coefficient of the highest power (x²) is one. Move the constant term to the right hand side. Take half of the coefficient of the middle term(x), square it, and add that value to both sides of the equation. Factor the perfect square trinomial. Take the square root of each side and solve.

With practice you will get the hang of it.

Let’s see how it works (click here):

Here’s another example (click here):

Until next time,

Michal

An equation in which one side is a perfect square trinomial can be easily solved by taking the square root of each side. Easy is good, so we basically want to force the quadratic equation into the form (x+a)²=x²+2ax+a².

All it takes is making sure that the coefficient of the highest power (x²) is one. Move the constant term to the right hand side. Take half of the coefficient of the middle term(x), square it, and add that value to both sides of the equation. Factor the perfect square trinomial. Take the square root of each side and solve.

With practice you will get the hang of it.

Let’s see how it works (click here):

Here’s another example (click here):

Until next time,

Michal

A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c are constants.

We will look at three different methods for solving quadratics: factorization, completing the square, and the quadratic formula. Symbolab is using all three methods, simply click the [+] next to the method of your choice to see the detailed solution steps.

Here’s how it looks like (click here to see on symbolab):

First up is factorization.

How do you factorize a quadratic? Use the Symbolab solutions technique that we saw in the middle school math solutions edition (To access Symbolab Factoring Calculator click here).

The trick is to get the equation to the form (x-u)(x-v)=0, now we have to solve much simpler equations (any number multiplied by zero equals zero ).

Let’s see how it works (click here):

Here’s another example just to get the hang of it (click here):

Until next time,

Michal

We will look at three different methods for solving quadratics: factorization, completing the square, and the quadratic formula. Symbolab is using all three methods, simply click the [+] next to the method of your choice to see the detailed solution steps.

Here’s how it looks like (click here to see on symbolab):

First up is factorization.

How do you factorize a quadratic? Use the Symbolab solutions technique that we saw in the middle school math solutions edition (To access Symbolab Factoring Calculator click here).

The trick is to get the equation to the form (x-u)(x-v)=0, now we have to solve much simpler equations (any number multiplied by zero equals zero ).

Let’s see how it works (click here):

Here’s another example just to get the hang of it (click here):

Until next time,

Michal

Symbolab's "getting started" series is moving on to help you solve high school level algebra and calculus.

First up is solving high school level inequalities; that is quadratic inequalities and inequalities involving algebraic fractions.

This is a lot like solving simple inequalities (See Middle School Inequalities post here).

Just like with simple inequalities we start by solving the equation and same as before, we are looking for intervals rather than points. So what’s the catch? If you remember, we can’t simply multiply or divide by algebraic factors (multiplying or dividing by negative values reverse the inequality). Now the two intercepts split the x-axis into three intervals, we have to test the conditions for each. Finally we have to find the intervals that satisfy the condition.

Here's how Symbolab solves quadratic inequality, including the solution steps, (click here):

Here’s an example with algebraic fractions (click here):

Until next time,

Michal

First up is solving high school level inequalities; that is quadratic inequalities and inequalities involving algebraic fractions.

This is a lot like solving simple inequalities (See Middle School Inequalities post here).

Just like with simple inequalities we start by solving the equation and same as before, we are looking for intervals rather than points. So what’s the catch? If you remember, we can’t simply multiply or divide by algebraic factors (multiplying or dividing by negative values reverse the inequality). Now the two intercepts split the x-axis into three intervals, we have to test the conditions for each. Finally we have to find the intervals that satisfy the condition.

Here's how Symbolab solves quadratic inequality, including the solution steps, (click here):

Here’s an example with algebraic fractions (click here):

Until next time,

Michal

Ski Vacation? Nope, this is serious stuff; it’s about finding the slope of a line, finding the equation of a line with a given slope, finding the equation of the tangent line to a function, finding the distance between two points, finding the quadrant of a point, finding the midpoint of a line segment. That’s a lot of lines and points, and functions too.

Basically, you have to follow the rules, no tricks, just knowing how to use it all. We’ve added the line features to help you sort this through…step by step

Let’s start with a couple of examples.

Find the equation of a line given a point and slope m (click here):

All it takes is solving a linear equation to compute the y=intercept b

Let’s continue with the most challenging one, the tangent line. It is only challenging b/c it requires derivation, and it is using all the line features we’ve seen so far. So we have a function and a point, we have to start by finding the slope (that’s where we derivate), compute the slope at a given point, now we can find the tangent line ( slope + point = line)

Here is an example (click here):

You can check out more examples here.

Cheers,

Michal

Basically, you have to follow the rules, no tricks, just knowing how to use it all. We’ve added the line features to help you sort this through…step by step

Let’s start with a couple of examples.

Find the equation of a line given a point and slope m (click here):

All it takes is solving a linear equation to compute the y=intercept b

Let’s continue with the most challenging one, the tangent line. It is only challenging b/c it requires derivation, and it is using all the line features we’ve seen so far. So we have a function and a point, we have to start by finding the slope (that’s where we derivate), compute the slope at a given point, now we can find the tangent line ( slope + point = line)

Here is an example (click here):

You can check out more examples here.

Cheers,

Michal

Factoring is another very handy math skill where Symbolab can help you find the solution and also help you learn the solution steps.

Say you have a piece of algebra such as: x^2- 5x+6: the goal of factoring is to get this into a form that looks like (x+u)(x+v).

Why would you want to do that? to break the expression into more manageable chunks. It turns out to be a super-important step when we come to solve quadratic equations e.g. x ^2 - 5x+6=0 (click here if you want to see how it works ).

The key to factoring quadratic x^2 - 5x + 6 is to find u and v that factor to 6 and add up to -5.

Here's how the solution looks like (click here):

And here's one more, slightly trickier example (click here):

Feel free to play with more examples on Symbolab until you've properly got the hang of it.

Cheers,

Michal

Say you have a piece of algebra such as: x^2- 5x+6: the goal of factoring is to get this into a form that looks like (x+u)(x+v).

Why would you want to do that? to break the expression into more manageable chunks. It turns out to be a super-important step when we come to solve quadratic equations e.g. x ^2 - 5x+6=0 (click here if you want to see how it works ).

The key to factoring quadratic x^2 - 5x + 6 is to find u and v that factor to 6 and add up to -5.

Here's how the solution looks like (click here):

And here's one more, slightly trickier example (click here):

Feel free to play with more examples on Symbolab until you've properly got the hang of it.

Cheers,

Michal

Solving simultaneous equations is one small algebra step further on from simple equations. Symbolab math solutions feature can help with this too.

In simultaneous equations there are two or more unknowns. (In middle school you only need to know how to solve equations with two unknowns. In high school and college you may have to solve for three or more… but that's for later).

If you have two equations and two unknowns - call them x and y - then we're in business; Symbolab can help you find the solution.

Here's the trick - you use one of the two equations to express x in terms of y, putting the equation in the form x=... Then you substitute for x in terms of y in the second equation. That gives you one equation with one unknown, which we already know how to solve (link back to first blog post).

Here's an example (click here):

And here's another solution (click here):

If you want to take a peek at what's in store in high school math check out how to solve simultaneous equations with three unknowns using the same trick here.

Cheers,

Michal

In simultaneous equations there are two or more unknowns. (In middle school you only need to know how to solve equations with two unknowns. In high school and college you may have to solve for three or more… but that's for later).

If you have two equations and two unknowns - call them x and y - then we're in business; Symbolab can help you find the solution.

Here's the trick - you use one of the two equations to express x in terms of y, putting the equation in the form x=... Then you substitute for x in terms of y in the second equation. That gives you one equation with one unknown, which we already know how to solve (link back to first blog post).

Here's an example (click here):

And here's another solution (click here):

If you want to take a peek at what's in store in high school math check out how to solve simultaneous equations with three unknowns using the same trick here.

Cheers,

Michal

Next up in our Getting Started maths solutions series is help with another middle school algebra topic - solving simple inequalities.

For example, take an inequality with one unknown x. You have an expression on one side that you know is more or less than what's on the other side.

This is a lot like solving equations (check here). Here too, your goal is to get x by itself on one side and a number on the other.

You do this by adding, subtracting, multiplying or dividing both sides of the inequality. Remember, whatever you do to one side of the inequality, you have to do the same to the other side. Most important to note is that multiplication or division by a negative number reverses the inequality.

Here's how Symbolab solves it, including the solution steps to help you learn this skill for yourself (click here):

A bit more complicated inequalities might include both lower and upper bounds, or in some cases absolute values. Here you simply break the inequality into simple inequalities, solve each inequality and combine the solutions (some basic understanding of ‘and’, ‘or’ and ranges is required).

Here’s an example to demonstrate the steps (click here):

Until next time,

Michal

For example, take an inequality with one unknown x. You have an expression on one side that you know is more or less than what's on the other side.

This is a lot like solving equations (check here). Here too, your goal is to get x by itself on one side and a number on the other.

You do this by adding, subtracting, multiplying or dividing both sides of the inequality. Remember, whatever you do to one side of the inequality, you have to do the same to the other side. Most important to note is that multiplication or division by a negative number reverses the inequality.

Here's how Symbolab solves it, including the solution steps to help you learn this skill for yourself (click here):

A bit more complicated inequalities might include both lower and upper bounds, or in some cases absolute values. Here you simply break the inequality into simple inequalities, solve each inequality and combine the solutions (some basic understanding of ‘and’, ‘or’ and ranges is required).

Here’s an example to demonstrate the steps (click here):

Until next time,

Michal

Welcome to our new "Getting Started" math solutions series. Over the next few weeks, we'll be showing how Symbolab super-easy-to-use math solutions can help you solve math problems in algebra, calculus and more, for students from 6th grade through college and beyond.

As well as giving you the solution, Symbolab also shows you the solution steps, to help you learn the math skills you need.

Let's start with middle school algebra and simple equations. You have an equation with one unknown - call it x. The trick here to solving the equation is to end up with x on one side of the equation and a number on the other. You do this by adding, subtracting, multiplying or dividing both sides of the equation. Remember, whatever you do to one side of the equation, you have to do the same to the other side. (don’t worry, we’ll show you all the steps).

To enter an equation, just use your key pad or Symbolab's special key pad at the top of the screen.

Here's an example of how it works, (click here):

And here's another one which is slightly more complicated - here you have a fraction on one side of the equation; first step is to get rid of it by multiplying both sides of the equation by the number that's underneath the line in the fraction, (click here):

Try some of the other examples that Symbolab suggests on the side of the page until you get the hang of it!

Until next time,

Michal

As well as giving you the solution, Symbolab also shows you the solution steps, to help you learn the math skills you need.

Let's start with middle school algebra and simple equations. You have an equation with one unknown - call it x. The trick here to solving the equation is to end up with x on one side of the equation and a number on the other. You do this by adding, subtracting, multiplying or dividing both sides of the equation. Remember, whatever you do to one side of the equation, you have to do the same to the other side. (don’t worry, we’ll show you all the steps).

To enter an equation, just use your key pad or Symbolab's special key pad at the top of the screen.

Here's an example of how it works, (click here):

And here's another one which is slightly more complicated - here you have a fraction on one side of the equation; first step is to get rid of it by multiplying both sides of the equation by the number that's underneath the line in the fraction, (click here):

Try some of the other examples that Symbolab suggests on the side of the page until you get the hang of it!

Until next time,

Michal

When solving trigonometric equations you should be familiar with algebraic manipulations, basic geometry, triangles and more triangles, the six trigonometric ratios, trigonometric identities, trigonometric manipulations, radians, degrees and the unit circle. That’s quite a lot to remember.

To complicate things if there is one solution then there are an infinite number of solutions; trigonometric functions are periodic.

Trying to simplify things, the goal is to isolate the trigonometric function (using the scary list from above). To simplify even more, we are now solving trigonometric equations step by step. Let’s take a look at a few examples:

Example using trigonometric identities (click here):

Example using basic Algebraic manipulations (click here):

Stay tuned, we have more trig coming your way

Cheers,

Michal

To complicate things if there is one solution then there are an infinite number of solutions; trigonometric functions are periodic.

Trying to simplify things, the goal is to isolate the trigonometric function (using the scary list from above). To simplify even more, we are now solving trigonometric equations step by step. Let’s take a look at a few examples:

Example using trigonometric identities (click here):

Example using basic Algebraic manipulations (click here):

Stay tuned, we have more trig coming your way

Cheers,

Michal

To prove a trigonometric identity you have to show that one side of the equation can be transformed into the other side of the equation. This usually requires turning everything into sines and cosines, some algebraic manipulations (join fractions, expand, factor, etc.) and use of fundamental trigonometric identities. Sounds familiar? Not really, it is very different than solving equations. You have to work on one side at a time, so algebraic properties that work on both sides of the equation can’t be used. Complicated? Yes… but the good news about proofs is that you always know where you are going…

…and the real good news is that we are now doing it for you! Just say the magic word (Prove, not please), type in the trig identity, press Go, and wow, a step by step poof coming your way.

Let’s start with a few examples. With time (and practice) you’ll learn the best tricks to use and the identities that are most helpful (the Pythagorean identity is a favorite)

Example expressing with sine and cosine, Join, Factor and Pythagorean identity (click here)

Example using fraction multiplication trick (click here)

Enjoy the rest of the summer!

Michal

…and the real good news is that we are now doing it for you! Just say the magic word (Prove, not please), type in the trig identity, press Go, and wow, a step by step poof coming your way.

Let’s start with a few examples. With time (and practice) you’ll learn the best tricks to use and the identities that are most helpful (the Pythagorean identity is a favorite)

Example expressing with sine and cosine, Join, Factor and Pythagorean identity (click here)

Example using fraction multiplication trick (click here)

Enjoy the rest of the summer!

Michal

Adding or subtracting fractions, there is only one thing to remember, the denominators must be the same. Not so fast, there is more to it, like how to find the Least Common Denominator (the smallest number that all denominators can divide into) or how to cancel algebraic fractions.

Symbolab solves algebraic fractions step-by-step. Most important, you will not only get solution steps, but also intermediate steps for the algebraic manipulations such as cancel, expand, factor, etc.

Let’s start by adding fractions (click here):

Cancel example (click here):

Now let’s put it all together and try to solve equations with rational expressions

(click here):

Start your fractions here.

Cheers,

Michal

Symbolab solves algebraic fractions step-by-step. Most important, you will not only get solution steps, but also intermediate steps for the algebraic manipulations such as cancel, expand, factor, etc.

Let’s start by adding fractions (click here):

Cancel example (click here):

Now let’s put it all together and try to solve equations with rational expressions

(click here):

Start your fractions here.

Cheers,

Michal

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