How to get the results of this logarithmic equation? How to solve this for $x$: 
$$\log_x(x^3+1)\cdot\log_{x+1}(x)>2$$
I have tried to get the same exponent by getting the second 
multiplier to reciprocal and tried to simplify $(x^3+1)$.
 A: We have that
$$\begin{align}
\log_{x+1}x&=\frac{\log_xx}{\log_x(x+1)}\\\\
&=\frac{1}{\log_x(x+1)} \tag 1
\end{align}$$
and
$$\log_x(x^3+1)=\log_x (x+1)+\log_x(x^2-x+1) \tag 2$$
Using $(1)$ and $(2)$ reveals that 
$$\begin{align}
\log_x(x^3+1)\log_{x+1}x&=\left(\log_x (x+1)+\log_x(x^2-x+1)\right)\frac{1}{\log_x(x+1)}\\\\
&=1+\frac{\log_x(x^2-x+1)}{\log_x(x+1)} \tag 3
\end{align}$$
We note that if the right-hand side of $(3)$ is to be greater than $2$, we must have $\frac{\log_x(x^2-x+1)}{\log_x(x+1)}>1\implies \frac{x^2-x+1}{x+1}>1\implies x(x-2)>0\implies x>2 \,\,\text{for real-valued solutions}$
Thus, we have that 
$$\bbox[5px,border:2px solid #C0A000]{\log_x(x^3+1)\log_{x+1}x>2\,\,\text{for}\,\,x>2}$$
A: Keep in mind $(\log_a b)(\log_c a) = \log_c b$.  
Setting $a = x, b = x^3 + 1$, and $c = x + 1$, we have
$\log_x (x^3 + 1)\cdot \log_{x+1}x = \log_{x + 1}(x^3 + 1)$.
So your inequality is equivalent to
$\log_{x + 1}(x^3 + 1) > 2 
\\  x^3 + 1 > (x + 1)^2 
\\  x^3 + 1 > x^2 + 2x + 1
\\ x^3 - x^2 - 2x > 0
\\ x(x + 1)(x - 2) > 0$
Then $-1 < x < 0$ or $x > 2$. One case is ruled out.
