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

Step-by-step Solution

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Final answer to the problem

$x=3$
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Step-by-step Solution

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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

Apply the formula: $\log_{b}\left(x\right)-\log_{b}\left(y\right)$$=\log_{b}\left(\frac{x}{y}\right)$, where $b=10$, $x=x^2$ and $y=x+6$

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

Apply the formula: $\log_{b}\left(x\right)=a$$\to \log_{b}\left(x\right)=\log_{b}\left(b^a\right)$, where $a=0$, $b=10$, $x=\frac{x^2}{x+6}$ and $b,x=10,\frac{x^2}{x+6}$

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

Apply the formula: $\log_{a}\left(x\right)=\log_{a}\left(y\right)$$\to x=y$, where $a=10$, $x=\frac{x^2}{x+6}$ and $y=1$

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

Apply the formula: $\frac{a}{b}=c$$\to a=cb$, where $a=x^2$, $b=x+6$ and $c=1$

$x^2=x+6$
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

Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values

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

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

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

Solve the equation ($1$)

$x+2=0$
11

Apply the formula: $x+a=b$$\to x+a-a=b-a$, where $a=2$, $b=0$, $x+a=b=x+2=0$ and $x+a=x+2$

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

Apply the formula: $x+a+c=b+f$$\to x=b-a$, where $a=2$, $b=0$, $c=-2$ and $f=-2$

$x=-2$
13

Solve the equation ($2$)

$x-3=0$
14

Apply the formula: $x+a=b$$\to x+a-a=b-a$, where $a=-3$, $b=0$, $x+a=b=x-3=0$ and $x+a=x-3$

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

Apply the formula: $x+a+c=b+f$$\to x=b-a$, where $a=-3$, $b=0$, $c=3$ and $f=3$

$x=3$
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

$x=3$

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Plotting: $2\log \left(x\right)-\log \left(x+6\right)$

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7
8
9
0
a
b
c
d
f
g
m
n
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

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