A lot of logarithms!!

We should use here the formula $\log (\frac{m}{n}) = \log m - \log n$ and $m \log n = \log m^n$ . For the terms $\log 178, \log \sqrt{81(2^x+2\times 3^x)}, x \log 3$ to be in AP, then the difference of the consecutive terms should be same i.e,

$\large \log \sqrt{81(2^x+2\times 3^x)}-\log 178 = x \log 3-\log \sqrt{81(2^x+2\times 3^x)}$

$\implies \large{ \log (\frac{\sqrt{81(2^x+2\times 3^x)}}{178})= \log 3^x-\log \sqrt{81(2^x+2\times 3^x)}}$

$\implies \large{ \log (\frac{\sqrt{81(2^x+2\times 3^x)}}{178})= \large \log (\frac{3^x}{\sqrt{81(2^x+2\times 3^x)}})}$

Applying antilog, we get---

$\large{\frac{\sqrt{81(2^x+2\times 3^x)}}{178}=\frac{3^x}{\sqrt{81(2^x+2\times 3^x)}}}$

$\implies 178\times 3^x=81(2^x+2\times 3^x)$

$\implies 178\times 3^x=81\times 2^x + 162\times 3^x$

$\implies 16\times 3^x=81\times 2^x$

$\implies \large{\frac{2^x}{3^x}=\frac{16}{81}=\frac{2^4}{3^4}}$

$\implies \large{(\frac{2}{3})^x=(\frac{2}{3})^4}$

Comparing both sides, we get ----

$x=\boxed{4}$

If the numbers $a,b$ and $c$ are in arithmetic progression, it follows that $b-a=c-b$ . Also, a logarithm property states that $\log a - \log b = \log{\frac{a}{b}}$ .

It all comes down to:

$\log{\frac{\sqrt{81(2^x+2\times3^x)}}{178}} = \log{\frac{3^x}{\sqrt{81(2^x+2\times3^x)}}} \Rightarrow \frac{\sqrt{81(2^x+2\times3^x)}}{178} = \frac{3^x}{\sqrt{81(2^x+2\times3^x)}}$

$\Rightarrow 178 \times 3^x = 81 \times 2^x +162 \times 3^x \Rightarrow 16 \times 3^x = 81 \times 2^x$

$\Rightarrow \frac{3^x}{81} = \frac{2^x}{16} \Rightarrow 3^{x-4} = 2^{x-4}$ .

A generic form of this last equation is $a^c = b^c$ , where $\; a$ and $\; b$ are constants and $\; c$ is the variable. For $a \neq b$ , it will only be true for $c=0$ .

Thus, $x-4=0 \Rightarrow \boxed{x = 4.}$