$$2-\log_3(x-7)=\log_{\frac13}(2x)$$
$$2-\frac{\log(x-7)}{\log3}=\frac{\log(2x)}{\log{\frac13}}$$
$$2-\frac{\log(x-7)}{\log3}=\frac{\log(2x)}{-\log{3}}$$
$$2\log(3)-\log(x-7)=-\log(2x)$$
$$\log{\frac{9}{x-7}}=\log{\frac{1}{2x}}$$
$$\frac{9}{x-7}=\frac{1}{2x}$$
$$18x=x-7$$
$$x=\frac{-7}{17}$$
But for logarithm to be defined, we must have $x-7\gt0$ and $2x\gt0$. Thus, there is no solution.
Start with
$\dfrac{a+b}{2}
\ge \sqrt{ab}
$.
To prove this,
write it as
$\dfrac{a-2\sqrt{ab}+b}{2}
\ge 0
$,
and the left side is
$\dfrac{(\sqrt{a}-\sqrt{b})^2}{2}
\ge 0
$.
Then,
$\begin{array}\\
\dfrac{a+b+c+d}{4}
&=\dfrac{a+b}{4}+\dfrac{c+d}{4}\\
&=\dfrac{\dfrac{a+b}{2}}{2}+\dfrac{\dfrac{c+d}{2}}{2}\\
&\ge\dfrac{\sqrt{ab}}{2}+\dfrac{\sqrt{cd}}{2}\\
&=\dfrac{\sqrt{ab}+\sqrt{cd}}{2}\\
&\ge\sqrt{\sqrt{ab}\sqrt{cd}}\\
&=\sqrt{\sqrt{abcd}}\\
&=\sqrt[4]{abcd}\\
\end{array}
$
By induction on $n$,
with this technique
you can show that
$\dfrac{\sum_{k=1}^{2^n}a_k}{2^n}
\ge \sqrt[2^n]{\prod_{k=1}^n a_k}
$.
To show this is true
for any $m < 2^n$,
let
$a_j
=\dfrac{\sum_{k=1}^m a_k}{m}
$
for $j \gt m$
and see what happens.
As a matter of fact,
this was Cauchy's
original proof.
Here's the details (added later).
The left side is,
letting
$a = \dfrac{\sum_{k=1}^m a_k}{m}
$,
$\begin{array}\\
\dfrac{\sum_{k=1}^{2^n}a_k}{2^n}
&=\dfrac{\sum_{k=1}^{m}a_k}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a_k}{2^n}\\
&=\dfrac{\sum_{k=1}^{m}a_k}{m}\dfrac{m}{2^n}+\dfrac{\sum_{k=m+1}^{2^n}a}{2^n}\\
&=\dfrac{am}{2^n}+\dfrac{(2^n-m)a}{2^n}\\
&=\dfrac{am}{2^n}+\dfrac{2^na}{2^n}-\dfrac{ma}{2^n}\\
&= a\\
&=\dfrac{\sum_{j=1}^ma_j}{m}\\
\end{array}
$
Similarly,
the right side is,
letting
$a_j
=b
=\left(\prod_{k=1}^{m} a_k\right)^{1/m}
$
for $j > m$,
$\begin{array}\\
\sqrt[2^n]{\prod_{k=1}^{2^n} a_k}
&=\left(\prod_{k=1}^{2^n} a_k\right)^{1/2^n}\\
&=\left(\prod_{k=1}^{m} a_k\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\
&=\left(\prod_{k=1}^{m} a_k\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} a_k\right)^{1/2^n}\\
&=\left(b^m\right)^{1/2^n}\left(\prod_{k=m+1}^{2^n} b\right)^{1/2^n}\\
&=b^{m/2^n}\left(b^{2^n-m}\right)^{1/2^n}\\
&=b^{m/2^n}b^{(2^n-m)/2^n}\\
&=b\\
&=\left(\prod_{k=1}^{m} a_k\right)^{1/m}\\
\end{array}
$
Therefore
$a \ge b$
or
$\dfrac{\sum_{j=1}^ma_j}{m}
\ge \left(\prod_{k=1}^{m} a_k\right)^{1/m}
$.
Best Answer
Consider the left hand side:
$$\log_2{x}+\log_3{x}+\log_5{x} = \frac{\log_{30}{x}}{\log_{30}{2}}+\frac{\log_{30}{x}}{\log_{30}{3}}+\frac{\log_{30}{x}}{\log_{30}{5}}.$$
Factor out $\log_{30}{x}$:
$$\log_2{x}+\log_3{x}+\log_5{x} = \log_{30}x\left(\frac{1}{\log_{30}{2}}+\frac{1}{\log_{30}{5}}+\frac{1}{\log_{30}{5}}\right).$$
Now use the fact that $\frac{1}{\log_b{a}} = \log_a{b}$:
$$\log_2{x}+\log_3{x}+\log_5{x} = \log_{30}{x}\left(\log_2{30}+\log_3{30}+\log_5{30}\right).$$
So now all we need to show is
$$\log_2{30}+\log_3{30}+\log_5{30}>9.$$
$\log_2{16}=4$, and since $\log_2$ is a strictly increasing function, this means $\log_2{30}>(\log_2{16}=)4$. Applying the same logic to the other two $\log$s, we get that $\log_3{30}>3$ and $\log_5{30}>2$, hence
$$\log_2{30}+\log_3{30}+\log_5{30}>4+3+2=9$$
and we are done