Help with Logarithms that have uneven bases?
6^x = 24
8^(x - 2) = 3
5^(x + 5) = 9^x
3^(x - 4) = 7^(x + 9)
I have a test I need to take on these tomorrow and I'm REALLY freaking over because I just don't understand them ); If you could take me through the steps of each ones (or just go through one and provide an answer key so I can check my answers against it) that would be awesome! Thanks!

Question

Help with Logarithms that have uneven bases?
6^x = 24
8^(x - 2) = 3
5^(x + 5) = 9^x
3^(x - 4) = 7^(x + 9)
I have a test I need to take on these tomorrow and I'm REALLY freaking over because I just don't understand them ); If you could take me through the steps of each ones (or just go through one and provide an answer key so I can check my answers against it) that would be awesome! Thanks!

anonymous · Answer

$$
6^x=24\
x=\log_624\
x=\frac{\ln24}{\ln6}
$$
$$
8^{x-2}=3\
x-2=\log_83
$$Do you follow?

anonymous · Answer

I think so ); But ... Gah, probably not. I'm so incredibly bad at these things. Sorry for taking up your time like this! So, with 6^x = 24, you just switch it to a log with a base of 6 and keep 24 at the end like that? So, if it were 3^x = 5, you'd just put log3 5?

anonymous · Answer

Depends on what your teacher wants, you can simplify it a little bit more if you really need to but the answer is irrational so no matter what it'll have some $\ln$ terms. You just remember what logarithm means, $\log_ab=x$ is the exact same thing as saying $a^x=b$.