Question

log5^x-log5^17=4

Answer 1

(x-17)ln5=4

Answer 2

how? (x-17)log5=4?

Answer 3

anytime you have a log you can take the exponent and move it outside of the log. So lnx^2=2lnx

Answer 4

I'm not sure what you're trying to ask. Is it this?
\[\log_5(x)-\log_5(17)=4\]

How about this?
\[(\log_a5)^x-(\log_a5)^{17}=4\]

or this maybe?
\[\log_{a}(5^{x})-\log_{a}(5^{17})=4\]

The last of these is what Rouault has solved for you.

Answer 5

\(\log_5(x)-log_5(17)=4\\\log_5(x)=4+\log_5(17)\\x=5^{4+\log_5(17)}=5^45^{\log_5(17)}=625(17)=10625\)

Answer 6

that is how to solve the last one, im not sure what @rouault did...

Answer 7

the right one is @zzruck3r how do it become x = 5^4+log5(17)?

Answer 8

\(\log_5(x)=4+\log_5(17)\\5^{\log_5(x)}=5^{4+\log_5(17)}\\x=5^{4+\log_5(17)}\)

Answer 9

okayy thank you i got it!

Answer 10

np