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**Hint:**For simplifying the original equation , firstly used logarithm property

${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$ then take base two exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the given equation .

**Formula used:**

We used logarithm properties i.e.,

${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$

And

${b^{{{\log }_b}a}} = a$ , the logarithm function says $\log x$ is only defined when $x$ is greater than zero.

**Complete solution step by step:**

It is given that ,

${\log _2}x = {\log _4}(x + 6)$ ,

We have to solve for $x$ .

Now using logarithm property ${\log _b}a = \dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$ ,

We will get,

${\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{{{{\log }_2}(4)}}$

Now , simplify the equation , we will get the following result ,

$ \Rightarrow {\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{{{{\log }_2}{{(2)}^2}}}$ (using $\log {a^b}

= b\log a$)

$ \Rightarrow {\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{{2{{\log }_2}(2)}}$

Using ${\log _a}a = 1$ ,

$ \Rightarrow {\log _2}x = \dfrac{{{{\log }_2}(x + 6)}}{2}$

$

\Rightarrow 2{\log _2}x = {\log _2}(x + 6) \\

\Rightarrow {\log _2}{(x)^2} = {\log _2}(x + 6) \\

$ (using $\log {a^b} = b\log a$)

Now , by assuming the base of the logarithm to be $2$ ,then take the base $2$ exponential of both sides of the equation, we will get the following result ,

For equation one ,

${2^{{{\log }_2}\left( {{x^2}} \right)}} = {2^{{{\log }_2}(x + 6)}}$

By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get the following result ,

${x^2} = x + 6$

Or

${x^2} - x - 6 = 0$

$

\Rightarrow {x^2} - 3x + 2x - 6 = 0 \\

\Rightarrow x(x - 3) + 2(x - 3) = 0 \\

\Rightarrow (x + 2)(x - 3) = 0 \\

\Rightarrow x = - 2 \\

and \\

x = 3 \\

$

Now recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero.

Therefore, in our original equation ${\log _2}x = {\log _4}(x + 6)$ ,

Here,

$

(x) > 0 \\

and \\

(x + 6) > 0 \\

$ ,

When $x = - 2$ , it is less than zero.

Therefore , we reject $x = - 2$ .

And when $x = 3$ , it is greater than zero .

**Therefore, we have our solutions i.e., $x = 3$ .**

**Note:**The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$ is greater than zero. While performing logarithm properties we have to remember certain conditions , our end result must satisfy the domain of that logarithm .

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