Question

log 4 32

Answer 1

type it in your calculator

Answer 2

Use the change of base formula to get it into a base ten so you can type it into your calculator

Answer 3

that's where I get confused is the change of base. idk how to do that

Answer 4

it should be in your textbook

Answer 5

I don't have a text book.... :(

Answer 6

i do online and i don't understand how it words it

Answer 7

\[\log b (x) = \frac{ \log x }{ \log b }\]

Answer 8

what is x and what is b?

Answer 9

\[\log_4 32=\frac{\log_232}{\log_24}=\frac{\log_22^5}{\log_22^2}=\]

Answer 10

thank you @UnkleRhaukus
that helped so much!

Answer 11

what did you get for the final result @soccerbabe239?

Answer 12

i got 8

Answer 13

hmm, in that case something went wrong

Answer 14

what do you mean? D:

Answer 15

\[\log_4 32\]changing the base to base two \[=\frac{\log_232}{\log_24}\] rewriting the numbers as powers \[=\frac{\log_22^5}{\log_22^2}\]

now use this logarithm rule
\[\log_bx^n=n\log_b x\]
and also remember that
\[\log_bb=1\]

Answer 16

okay... im back to being confused hah!

Answer 17

first lets look at the numerator first
\[\log_22^5\]
do you see how we got this?

can you see what to do next?

Answer 18

thanks for the help but im kinda hopeless with this haha im just going to guess

Answer 19

using the rule\[\boxed{\log_bx^n=n\log_bx}\]
we see that \[\log_22^5=5\log_22\]

Answer 20

now using \[\boxed{\log_bb=1}\]
we see \[5\log_22=5\times1=5\]

Answer 21

repeat this process for the denominator \(\log_22^2\)