Is there a way to find x when x is an exponent? I'm kind of stuck on how to solve the following.
$10^x = 5^9$
Is there a method or a simple trick to find what is x?
 A: That's precisely what logarithms are for. In this case, the logarithm with base 10 (one of the two most common logarithms).
$$10^x=y$$
take the logarithm on both sides:
$$x=\log_{10}y$$
Using the properties of logarithm, your case gives $x=\log_{10}5^9=9\log_{10}5$.
A: Yes. Apply $\log_{10}$ in both sides of the equation. We get $\log_{10} 10^x = \log_{10} 5^9$, then $x \log_{10} 10 = 9 \log_{10} 5 $, and finally $x = 9 \log_{10} 5 $ using the properties of the logarithm. Always apply the logarithm with the basis on which $x$ is the exponent. In the example, I used $\log_{10}$ because we had $10^x$. Try to solve $3^x = 7^8$ to see if you understood the idea (:
A: Applying the logarithm is a valid way to do it, but it's basically just saying "there is a technical term for this". Just giving $x$ the name $\log_{10} y$ might not be helpful depending on what you need to do with $x$. Here's how you can go a little bit further.
First, $x$ has to be positive since $10$ raised to a negative power will be less than $1$.
Simple number theory shows $x$ can't be an integer, since any integer power of $10$ is even, and $5^9$ is odd.
If we want $x$ to be a rational number, then we have $\sqrt[b]{10}^a=5^9$, so a root of $10$ is equal to a root of $5^9$. That is, for some $r$, we have $r^b=10$ and $r^a=5^9$, so $r^{ab}=10^a=(5^9)^b$. Again, a power of $10$ can't be equal to a power of $5^9$ because a power of $10$ is always even but a power of $5^9$ is always odd. So $x$ is not rational.
But, we know that $x$ exists (and is unique) as a real number because $10^x$ is continuous, so we can use the intermediate value theorem. And indeed, the standard name for $x$ is $\log_{10} 5^9$.
You can get quick rational approximations for $x$ by realizing that $10^{k/n}>5^9$ is equivalent to $10^k>(5^{9})^n$. Just set $n$ to be a very large integer and then test successive values of $k$. For example, set $n=1000$. With a powerful calculator, you can show that $10^{6290}<5^{9n}$, but $10^{6291}>5^{9n}$. So $6.290<x<6.291$.
