Solve $(x+1)3^x > 3^{x+1}$ I'm only at a college algebra level (pre-calculus) and am having a hard time knowing what steps to perform to solve this as x is both an exponent and at the base level.  I put this into a calculator and got x = 2 but I'd like to know how to perform the steps.  I know logs are involved but haven't been able to figure it out.  Here is the problem again $(x + 1)3^x > 3^{x+1}$  Any help is greatly appreciated :)
 A: HINT:
$3^x>0$ for all $x$. Therefore, you can divide both sides by $3^x$ without changing the inequality. Logs should not be involved in simplifying this inequality.
Also, $3^{x+1} = 3^x\cdot 3^1$
A: $$(x+1)3^x>3^x\cdot3$$
dividing by $3^x$ you get:
$$(x+1)\gt 3$$ for $x\gt 2$
A: $(x+1)3^x> 3^{x+1}=3\cdot3^x$.
By subtracting $3\cdot3^x$ from both sides we get
$(x-2)3^x>0$
As $3^x$ is a positive real number for every real-valued $x$, we are able to divide both side by $3^x$ leaving us with $x-2>0$ which has the solution of $x>2$.
A: Taking from @Vanwij's approach:
We have \begin{align}(x+1)3^x&>3^{x+1}\\
(x+1)3^x-3^{x+1}&>0\\
3^x[(x+1)-3]&>0\\
3^x(x-2)&>0
\end{align}
Let $y=3^x(x-2)$.  The $x$-intercept of the function $y$ is $x=2$.  As such, solving using an interval table, we have
\begin{array}{|c|c|c|c|}
\hline
& 3^x & x-2 & \text{Sign of}\ y \\
\hline
x<2 & + & - & - \\
\hline
x=2 & + & 0 & 0 \\
\hline
x>2 & + & + & +\\
\hline
\end{array}
Since we are looking for $y>0$, the only interval that satisfies the inequality is $\boxed{x>2}$.
