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# Logarithmic Equations Calculator

## Get detailed solutions to your math problems with our Logarithmic Equations step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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###  Difficult Problems

1

Here, we show you a step-by-step solved example of logarithmic equations. This solution was automatically generated by our smart calculator:

$2log\left(x\right)-log\left(x+6\right)=0$
2

Apply the formula: $a\log_{b}\left(x\right)$$=\log_{b}\left(x^a\right)$

$\log \left(x^2\right)-\log \left(x+6\right)=0$
3

The difference of two logarithms of equal base $b$ is equal to the logarithm of the quotient: $\log_b(x)-\log_b(y)=\log_b\left(\frac{x}{y}\right)$

$\log \left(\frac{x^2}{x+6}\right)=0$

Rewrite the number $0$ as a logarithm of base $10$

$\log \left(\frac{x^2}{x+6}\right)=\log \left(10^0\right)$

Any expression (except $0$ and $\infty$) to the power of $0$ is equal to $1$

$\log \left(\frac{x^2}{x+6}\right)=\log \left(1\right)$
4

Rewrite the number $0$ as a logarithm of base $10$

$\log \left(\frac{x^2}{x+6}\right)=\log \left(1\right)$
5

For two logarithms of the same base to be equal, their arguments must be equal. In other words, if $\log(a)=\log(b)$ then $a$ must equal $b$

$\frac{x^2}{x+6}=1$
6

Multiply both sides of the equation by $x+6$

$x^2=x+6$
7

Move everything to the left hand side of the equation

$x^2-x-6=0$
8

Factor the trinomial $x^2-x-6$ finding two numbers that multiply to form $-6$ and added form $-1$

$\begin{matrix}\left(2\right)\left(-3\right)=-6\\ \left(2\right)+\left(-3\right)=-1\end{matrix}$
9

Thus

$\left(x+2\right)\left(x-3\right)=0$
10

Break the equation in $2$ factors and set each equal to zero, to obtain

$x+2=0,\:x-3=0$
11

Solve the equation ($1$)

$x+2=0$
12

We need to isolate the dependent variable , we can do that by simultaneously subtracting $2$ from both sides of the equation

$x+2-2=0-2$

Canceling terms on both sides

$x=0-2$

$x+0=x$, where $x$ is any expression

$x=-2$
13

Canceling terms on both sides

$x=-2$
14

Solve the equation ($2$)

$x-3=0$
15

We need to isolate the dependent variable , we can do that by simultaneously subtracting $-3$ from both sides of the equation

$x-3+3=0+3$

Canceling terms on both sides

$x=0+3$

$x+0=x$, where $x$ is any expression

$x=3$
16

Canceling terms on both sides

$x=3$
17

Combining all solutions, the $2$ solutions of the equation are

$x=-2,\:x=3$

Verify that the solutions obtained are valid in the initial equation

18

The valid solutions to the logarithmic equation are the ones that, when replaced in the original equation, don't result in any logarithm of negative numbers or zero, since in those cases the logarithm does not exist

$x=3$

##  Final answer to the problem

$x=3$

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