# 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 :)

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, $$3^{x+1} = 3^x\cdot 3^1$$

$$(x+1)3^x>3^x\cdot3$$ dividing by $$3^x$$ you get: $$(x+1)\gt 3$$ for $$x\gt 2$$

$$(x+1)3^x> 3^{x+1}=3\cdot3^x$$.

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 $$x>2$$.

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 $$\boxed{x>2}$$.