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$3^{x+1} = 3000$

How do I solve this? I know we use logarithms but I do not remember how to solve this kind of problem. I am guessing that I need to change the problem into log form. but how?

$\log_3{(x+1)} = 3 + \log_{10} 3$

what do I do next?

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    If you want to take base-3 logarithms, you don't get $\log_3(x+1)=3+\log_{10}3$. You get $x+1=\log_3 3000$ $=\log_3 3+\log_3 1000 = 1+\log_3 1000$.2012-12-07

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We interpret your question as asking about $3^{x+1}=3000$. If it is about $3^x+1=3000$, rewrite as $3^x=2999$ and use the same procedure as the one below.

Take the logarithm of both sides, any base you like. I suggest base $10$ or base $e$ ($\n$), because these are easiest to find with a scientific calculator. We get $(x+1)\log(3)=\log(3000).$ So $x+1=\frac{\log(3000)}{\log(3)}.$ Calculate.

Note: We used the important fact that $\log(a^b)=b\log a$.

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    Doesn't make any difference, let $b=x+1$, $a=3$. The log of $3^{x+1}$ is $b\log a$, that is, $(x+1)\log 3$.2012-12-07
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If you can get an equation into the form $a^{f(x)}=b$ for some $a,b>0$ and some function $f(x)$, then you may equivalently write $f(x)=\log_a b,$ or alternately, $f(x)=\frac{\log b}{\log a}.$ At that point, if $f(x)$ is a linear or quadratic function, you should hopefully know how to solve the resulting equation.