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## Resources to Help You Solve Math Equations

Whether you love math or suffer through every single problem, there are plenty of resources to help you solve math equations. Skip the tutor and log on to load these awesome websites for a fantastic free equation solver or simply to find answers for solving equations on the Internet.

## Stand By for Automatic Math Solutions at Quick Math

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## Modern Math Answers Come From Mathway

Mathway offers a free equation solver that sifts through your toughest math problems — and makes math easy. Simply enter your math problem into the Mathway calculator, and choose what you’d like the math management program to do with the problem. Pick from math solutions that include graphing, simplifying, finding a slope or solving for a y-intercept by scrolling through the Mathway drop-down menu. Use the answers for solving equations to explore different types of solutions, or set the calculator to offer the best solution for your particular math puzzle. Mathway offers the option to create an account, to sign in or sign up for additional features or to simply stick with the free equation solver.

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## Solving Log Equations with Exponentials

Using the Definition Using Exponentials Calculators & Etc.

The second type of log equation requires the use of The Relationship :

—The Relationship—

...........is equivalent to............ (means the exact same thing as)

log b ( y ) = x

In animated form, the two equations are related as shown below:

Content Continues Below

## MathHelp.com

Solving Logarithmic Equations

Note that the base in both the exponential form of the equation and the logarithmic form of the equation is " b ", but that the x and y switch sides when you switch between the two equations. If you can remember this — that whatever had been the argument of the log becomes the "equals" and whatever had been the "equals" becomes the exponent in the exponential, and vice versa — then you should not have too much trouble with solving log equations.

## Solve log 2 ( x ) = 4

Since this equation is in the form "log(of something) equals a number", rather than "log(of something) equals log(of something else)", I can solve the equation by using The Relationship:

log 2 ( x ) = 4

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## Solve log 2 (8) = x .

I can solve this by converting the logarithmic statement into its equivalent exponential form, using The Relationship:

log 2 (8) = x

But 8 = 2 3 , so I can equate powers of two:

Note that this could also have been solved by working directly from the definition of a logarithm.

What power, when put on " 2 ", would give you an 8 ? The power 3 , of course!

If you wanted to give yourself a lot of work, you could also do this one in your calculator, using the change-of-base formula:

log 2 (8) = ln(8) / ln(2)

Plug this into your calculator, and you'll get " 3 " as your answer. While this change-of-base technique is not particularly useful in this case, you can see that it does work. (Try it on your calculator, if you haven't already, so you're sure you know which keys to punch, and in which order.) You will need this technique in later problems.

I'm not saying that you'll necessarily want to solve equations using the change-of-base formula, or always by using the definition of logs, or any other particular method. But I am suggesting that you should make sure that you're comfortable with the various methods, and that you shouldn't panic if you and a friend used totally different methods for solving the same equation.

## Solve log 2 ( x ) + log 2 ( x − 2) = 3

I can't do anything with this equation yet, because I don't yet have it in the "log(of something) equals a number" form. So I'll need to use log rules to combine the two terms on the left-hand side of the equation:

log 2 ( x ) + log 2 ( x − 2) = 3

log 2 [( x )( x − 2)] = 3

log 2 ( x 2 − 2 x ) = 3

Now the equation is arranged in a useful way. At this point, I can use The Relationship to convert the log form of the equation to the corresponding exponential form, and then I can solve the result:

2 3 = x 2 − 2 x

8 = x 2 − 2 x

0 = x 2 − 2 x − 8

0 = ( x − 4)( x + 2)

x = 4, −2

But if x = −2 , then " log 2 ( x ) ", from the original logarithmic equation, will have a negative number for its argument (as will the term " log 2 ( x − 2)" ). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be x = −2 .

Then my solution is:

Keep in mind that you can always check your answers to any "solving" exercise by plugging those answers back into the original equation and checking that the solution "works". In this case, I'll plug my solution value into either side of the original equation, and verify that each side evaluates to the same number:

the left-hand side:

log 2 ( x ) + log 2 ( x − 2)

= log 2 (4) + log 2 (4 − 2)3

= log 2 (4) + log 2 (2)

= log 2 (2 2 ) + log 2 (2 1 )

= 2 + 1 = 3

## Solve log 2 (log 2 ( x )) = 1

This equation may look overly-complicated, but it's just another log equation. To solve this, I'll need to apply The Relationship twice. I start with the original equation and work with the "outer" log:

log 2 (log 2 ( x )) = 1

The Relationship converts the above to:

2 1 = log 2 ( x )

2 = log 2 ( x )

Now I'll apply The Relationship a second time:

Then the solution is:

## Solve log 2 ( x 2 ) = (log 2 ( x )) 2

First, I'll expand the square on the right-hand side to be the explicit product of two logs:

log 2 ( x 2 ) = [log 2 ( x )] 2

log 2 ( x 2 ) = [log 2 ( x )] [log 2 ( x )]

2·log 2 ( x ) = [log 2 ( x )] [log 2 ( x )]

Then I'll move that term from the left-hand side of the equation to the right-hand side:

0 = [log 2 ( x )] [log 2 ( x )] − 2·log 2 ( x )

This equation may look bad, but take a close look. It's nothing more than a factoring exercise at this point. So I'll factor, and then I'll solve the factors by using The Relationship:

0 = [log 2 ( x )] [log 2 ( x ) − 2]

log 2 ( x ) = 0 or log 2 ( x ) − 2 = 0

2 0 = x or log 2 ( x ) = 2

1 = x or 2 2 = x

1 = x or 4 = x

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## How to Solve Logarithms

Last Updated: May 10, 2023 References

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 208,971 times. Learn more...

Logarithms might be intimidating, but solving a logarithm is much simpler once you realize that logarithms are just another way to write out exponential equations. Once you rewrite the logarithm into a more familiar form, you should be able to solve it as you would solve any standard exponential equation.

## Before You Begin: Learn to Express a Logarithmic Equation Exponentially [1] X Research source [2] X Research source

- If and only if: b y = x
- b does not equal 1
- In the same equation, y is the exponent and x is the exponential expression that the logarithm is set equal to.

- Example: 1024 = ?

- This could also be written as: 4 5

- Example: 4 5 = 1024

## Method One: Solve for X

- log 3 ( x + 5) + 6 - 6 = 10 - 6
- log 3 ( x + 5) = 4

- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 4; b = 3; x = x + 5
- Rewrite the equation so that: b y = x
- 3 4 = x + 5

- 3 * 3 * 3 * 3 = x + 5
- 81 - 5 = x + 5 - 5

- Example: x = 76

## Method Two: Solve for X Using the Logarithmic Product Rule [3] X Research source [4] X Research source

- log b (m * n) = log b (m) + log b (n)

- log 4 (x + 6) + log 4 (x) = 2 - log 4 (x) + log 4 (x)
- log 4 (x + 6) + log 4 (x) = 2

- log 4 [(x + 6) * x] = 2
- log 4 (x 2 + 6x) = 2

- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 4 ; x = x 2 + 6x
- 4 2 = x 2 + 6x

- 4 * 4 = x 2 + 6x
- 16 = x 2 + 6x
- 16 - 16 = x 2 + 6x - 16
- 0 = x 2 + 6x - 16
- 0 = (x - 2) * (x + 8)
- x = 2; x = -8

- Example: x = 2
- Note that you cannot have a negative solution for a logarithm, so you can discard x - 8 as a solution.

## Method Three: Solve for X Using the Logarithmic Quotient Rule [5] X Research source

- log b (m / n) = log b (m) - log b (n)

- log 3 (x + 6) - log 3 (x - 2) = 2 + log 3 (x - 2) - log 3 (x - 2)
- log 3 (x + 6) - log 3 (x - 2) = 2

- log 3 [(x + 6) / (x - 2)] = 2

- Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 3; x = (x + 6) / (x - 2)
- 3 2 = (x + 6) / (x - 2)

- 3 * 3 = (x + 6) / (x - 2)
- 9 = (x + 6) / (x - 2)
- 9 * (x - 2) = [(x + 6) / (x - 2)] * (x - 2)
- 9x - 18 = x + 6
- 9x - x - 18 + 18 = x - x + 6 + 18
- 8x / 8 = 24 / 8

- Example: x = 3

## Community Q&A

## You Might Also Like

- ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut43_logfun.htm#logdef
- ↑ https://www.mathsisfun.com/algebra/logarithms.html
- ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm
- ↑ https://www.youtube.com/watch?v=fnhFneOz6n8
- ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm

## About This Article

To solve a logarithm, start by identifying the base, which is "b" in the equation, the exponent, which is "y," and the exponential expression, which is "x." Then, move the exponential expression to one side of the equation, and apply the exponent to the base by multiplying the base by itself the number of times indicated in the exponent. Finally, rewrite your final answer as an exponential expression. To learn how to solve for "x" in a logarithm, scroll down! Did this summary help you? Yes No

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- Solving Logarithmic Equations – Explanation & Examples

## How to solve equations with logarithms on one side?

How to solve equations with logarithms on both sides of the equation, practice questions, solving logarithmic equations – explanation & examples.

Before we can get into solving logarithmic equations, let’s first familiarize ourselves with the following rules of logarithms :

- The product rule:

The product rule states that the sum of two logarithms is equal to the product of the logarithms. The first law is represented as;

⟹ log b (x) + log b (y) = log b (xy)

- The quotient rule:

The difference of two logarithms x and y is equal to the ratio of the logarithms.

⟹ log b (x) – log b (y) = log (x/y)

- The power rule:

⟹ log b (x) n = n log b (x)

- Change of base rule.

⟹ log b x = (log a x) / (log a b)

- Identity rule

The logarithm of any positive number to the same base of that number is always 1. b 1 =b ⟹ log b (b)=1.

## How to Solve Logarithmic Equations?

An equation containing variables in the exponents is knowns as an exponential equation. In contrast, an equation that involves the logarithm of an expression containing a variable is referred to as a logarithmic equation.

The purpose of solving a logarithmic equation is to find the value of the unknown variable.

In this article, we will learn how to solve the general two types of logarithmic equations, namely:

- Equations containing logarithms on one side of the equation.
- Equations with logarithms on opposite sides of the equal to sign.

Equations with logarithms on one side take log b M = n ⇒ M = b n .

To solve this type of equations, here are the steps:

- Simplify the logarithmic equations by applying the appropriate laws of logarithms.
- Rewrite the logarithmic equation in exponential form.
- Now simplify the exponent and solve for the variable.
- Verify your answer by substituting it back in the logarithmic equation. You should note that the acceptable answer of a logarithmic equation only produces a positive argument.

Solve log 2 (5x + 7) = 5

Rewrite the equation to exponential form

logs 2 (5x + 7) = 5 ⇒ 2 5 = 5x + 7

⇒ 32 = 5x + 7

⇒ 5x = 32 – 7

Divide both sides by 5 to get

Solve for x in log (5x -11) = 2

Since the base of this equation is not given, we therefore assume the base of 10.

Now change the write the logarithm in exponential form.

⇒ 10 2 = 5x – 11

⇒ 100 = 5x -11

Hence, x = 111/5 is the answer.

Solve log 10 (2x + 1) = 3

Rewrite the equation in exponential form

log 10 (2x + 1) = 3n⇒ 2x + 1 = 10 3

⇒ 2x + 1 = 1000

On dividing both sides by 2, we get;

Verify your answer by substituting it in the original logarithmic equation;

⇒ log 10 (2 x 499.5 + 1) = log 10 (1000) = 3 since 10 3 = 1000

Evaluate ln (4x -1) = 3

Rewrite the equation in exponential form as;

ln (4x -1) = 3 ⇒ 4x – 3 =e 3

But as you know, e = 2.718281828

4x – 3 = (2.718281828) 3 = 20.085537

x = 5.271384

Solve the logarithmic equation log 2 (x +1) – log 2 (x – 4) = 3

First simplify the logarithms by applying the quotient rule as shown below.

log 2 (x +1) – log 2 (x – 4) = 3 ⇒ log 2 [(x + 1)/ (x – 4)] = 3

Now, rewrite the equation in exponential form

⇒2 3 = [(x + 1)/ (x – 4)]

⇒ 8 = [(x + 1)/ (x – 4)]

Cross multiply the equation

⇒ [(x + 1) = 8(x – 4)]

⇒ x + 1 = 8x -32

7x = 33 …… (Collecting the like terms)

Solve for x if log 4 (x) + log 4 (x -12) = 3

Simplify the logarithm by using the product rule as follows;

log 4 (x) + log 4 (x -12) = 3 ⇒ log 4 [(x) (x – 12)] = 3

⇒ log 4 (x 2 – 12x) = 3

Convert the equation in exponential form.

⇒ 4 3 = x 2 – 12x

⇒ 64 = x 2 – 12x

Since this is a quadratic equation, we therefore solve by factoring.

x 2 -12x – 64 ⇒ (x + 4) (x – 16) = 0

x = -4 or 16

When x = -4 is substituted in the original equation, we get a negative answer which is imaginary. Therefore, 16 is the only acceptable solution.

The equations with logarithms on both sides of the equal to sign take log M = log N, which is the same as M = N.

The procedure of solving equations with logarithms on both sides of the equal sign.

- If the logarithms have are a common base, simplify the problem and then rewrite it without logarithms.
- Simplify by collecting like terms and solve for the variable in the equation.
- Check your answer by plugging it back in the original equation. Remember that, an acceptable answer will produce a positive argument.

Solve log 6 (2x – 4) + log 6 ( 4) = log 6 (40)

First, simplify the logarithms.

log 6 (2x – 4) + log 6 (4) = log 6 (40) ⇒ log 6 [4(2x – 4)] = log 6 (40)

Now drop the logarithms

⇒ [4(2x – 4)] = (40)

⇒ 8x – 16 = 40

⇒ 8x = 40 + 16

Solve the logarithmic equation: log 7 (x – 2) + log 7 (x + 3) = log 7 14

Simplify the equation by applying the product rule.

Log 7 [(x – 2) (x + 3)] = log 7 14

Drop the logarithms.

⇒ [(x – 2) (x + 3)] = 14

Distribute the FOIL to get;

⇒ x 2 – x – 6 = 14

⇒ x 2 – x – 20 = 0

⇒ (x + 4) (x – 5) = 0

x = -4 or x = 5

when x = -5 and x = 5 are substituted in the original equation, they give a negative and positive argument respectively. Therefor, x = 5 is the only acceptable solution.

Solve log 3 x + log 3 (x + 3) = log 3 (2x + 6)

Given the equation; log 3 (x 2 + 3x) = log 3 (2x + 6), drop the logarithms to get; ⇒ x 2 + 3x = 2x + 6 ⇒ x 2 + 3x – 2x – 6 = 0 x 2 + x – 6 = 0……………… (Quadratic equation) Factor the quadratic equation to get;

(x – 2) (x + 3) = 0 x = 2 and x = -3

By verifying both values of x, we get x = 2 to be the correct answer.

Solve log 5 (30x – 10) – 2 = log 5 (x + 6)

log 5 (30x – 10) – 2 = log 5 (x + 6)

This equation can be rewritten as;

⇒ log 5 (30x – 10) – log 5 (x + 6) = 2

Simplify the logarithms

log 5 [(30x – 10)/ (x + 6)] = 2

Rewrite logarithm in exponential form.

⇒ 5 2 = [(30x – 10)/ (x + 6)]

⇒ 25 = [(30x – 10)/ (x + 6)]

On cross multiplying, we get;

⇒ 30x – 10 = 25 (x + 6)

⇒ 30x – 10 = 25x + 150

⇒ 30x – 25x = 150 + 10

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## Most Used Actions

Number line.

- \log _2(x+1)=\log _3(27)
- \ln (x+2)-\ln (x+1)=1
- \ln (x)+\ln (x-1)=\ln (3x+12)
- 4+\log _3(7x)=10
- \ln (10)-\ln (7-x)=\ln (x)
- \log _2(x^2-6x)=3+\log _2(1-x)
- How do you calculate logarithmic equations?
- To solve a logarithmic equations use the esxponents rules to isolate logarithmic expressions with the same base. Set the arguments equal to each other, solve the equation and check your answer.
- What is logarithm equation?
- A logarithmic equation is an equation that involves the logarithm of an expression containing a varaible.
- What are the 3 types of logarithms?
- The three types of logarithms are common logarithms (base 10), natural logarithms (base e), and logarithms with an arbitrary base.
- Is log10 and log the same?
- When there's no base on the log it means the common logarithm which is log base 10.
- What is the inverse of log in math?
- The inverse of a log function is an exponantial.

logarithmic-equation-calculator

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## Course: Algebra 2 > Unit 8

Solving exponential equations using logarithms: base-10.

- Solving exponential equations using logarithms
- Solving exponential equations using logarithms: base-2

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If we are given an equation with a logarithm of the same base on both sides we may simply equate the arguments. Step 1: Use the rules of exponents to

Section 6.4 : Solving Logarithm Equations · log4(x2−2x)=log4(5x−12) log 4 ( x 2 − 2 x ) = log 4 ( 5 x − 12 ) Solution · log(6x)−log(4−x)=log(3) log ( 6 x )

Solve log2(x) + log2(x − 2) = 3 · log2(x) + log2(x − 2) = 3. log2[(x)(x − 2)] = 3. log2(x2 − 2x) = 3. Now the equation is arranged in a useful way. · log2(x2

When looking at the problem equation, identify the base (b), exponent (y), and exponential expression (x). Example: 5 = log4(1024). b = 4; y = 5; x = 1024.

The equations with logarithms on both sides of the equal to sign take log M = log N, which is the same as M = N. The procedure of solving

Logarithms are mathematical functions that transpose an exponential function to make it easier to solve. When computing algebraic equations

log 2( x−6 x )=3+log 2(1− x ). Scroll to see more. Math Help Tailored For You. Word Problems.

You need to know the properties of logarithms in order to solve this problem.

When I computed the problem I got x > 5/6. I 1st divided both sides by 3, then set it up as e^4x + e^3 < e^1/3. which i