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# Solve the logarithmic equation $2\log \left(x\right)-\log \left(x+6\right)=0$

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##  Final answer to the problem

$x=3$
Got another answer? Verify it here!

##  Step-by-step Solution 

How should I solve this problem?

• Choose an option
• Solve for x
• Find the derivative using the definition
• Solve by quadratic formula (general formula)
• Simplify
• Find the integral
• Find the derivative
• Factor
• Factor by completing the square
• Find the roots
Can't find a method? Tell us so we can add it.
1

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

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

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

$\log \left(\frac{x^2}{x+6}\right)=0$
3

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

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

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 $b$

$\frac{x^2}{x+6}=1$
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5

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

$x^2=x+6$
6

Move everything to the left hand side of the equation

$x^2-x-6=0$
7

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

$\begin{matrix}\left(2\right)\left(-3\right)=-6\\ \left(2\right)+\left(-3\right)=-1\end{matrix}$
8

Thus

$\left(x+2\right)\left(x-3\right)=0$
9

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

$x+2=0,\:x-3=0$
10

Solve the equation ($1$)

$x+2=0$
11

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

$x+2-2=0-2$
12

Canceling terms on both sides

$x=-2$
13

Solve the equation ($2$)

$x-3=0$
14

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

$x-3+3=0+3$
15

Canceling terms on both sides

$x=3$
16

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

17

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

$x=3$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

SnapXam A2

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0
a
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u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

###  Main Topic: Logarithmic Equations

Are those equations in which the unknown variable appears within a logarithm.