# changing equation to logarithmic form and solving it

$$3^{x+1} = 3000$$

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?

• Is that supposed to be $3^{x+1}=3000$ or $3^x+1=3000$? Dec 6, 2012 at 23:38
• $3^{x+1}=3\cdot 10^3$, divide by 3 to get $3^x=10^3$, so you got $x=\log_3(10^3)=3\log_3(10)\backsimeq 2.09590$ and this is it. Dec 6, 2012 at 23:49
• It is 3^(x+1)=3000 Dec 6, 2012 at 23:58
• If you want to take base-3 logarithms, you don't get $\log_3(x+1)=3+\log_{10}3$. You get $x+1=\log_3 3000$ $=\log_3 3+\log_3 1000 = 1+\log_3 1000$. Dec 7, 2012 at 1:30

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 $\log(a^b)=b\log a$.
• Doesn't make any difference, let $b=x+1$, $a=3$. The log of $3^{x+1}$ is $b\log a$, that is, $(x+1)\log 3$. Dec 7, 2012 at 0:02
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.