$$3^{x+1} = 3000$$
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?
$$3^{x+1} = 3000$$
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?
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 $\log(a^b)=b\log a$.
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.