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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 :)

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

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

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

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 $x>2$.

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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 $\boxed{x>2}$.

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