<s>[INST] 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 $(x^3+1)$.
 [/INST] 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.
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