# Logarithmic non-integer fractional value

Would it be possible to show the breakdown of how $\log_4$ $32$ = $\frac{5}{2}?$

I have to come up w/ 11 more just like it & I'm not sure how you came up w/ the answer.

Thank you!

You could use the change of base formula, noting that $4$ and $32$ are each powers of 2...

$$\log_4(32) = \dfrac{\log_2(32)}{\log_2(4)} = 5/2$$

• Are there different examples using different powers? – Christina Maines Apr 22 '12 at 13:47
• Uh... yeah! Do you have a similar question (from the other $11$ problems), or is something not clear about this answer? – The Chaz 2.0 Apr 22 '12 at 13:54
• No, just a pain trying to come up w/ 11. It's for a presentation that I have to give tomorrow. We didn't really go over this stuff in class & the book isn't any help @ all. I'm trying to avoid coming up w/ the same answers but it's kind of hard not too. – Christina Maines Apr 22 '12 at 14:18
• You are trying to have logarithmic expressions that are not integer-valued? How about $$\log_b(b^{1.2387}) = 1.2387$$ – The Chaz 2.0 Apr 22 '12 at 14:28
• Non-integer fractional values – Christina Maines Apr 22 '12 at 15:09

Alternatively you could say: $$log_432=log_42^5=log_4(4^{\frac{1}{2})^5}=log_44^{\frac{5}{2}}=\frac{5}{2}log_44=\frac{5}{2}$$

• yes it is very good representation,+1 – dato datuashvili Apr 22 '12 at 13:40

please look at $log_4(32)$ ,here $4=2^2$ and because it is log function,we can factor out 2 as 1/2,and $32=2^5$ ,for this ,we simple take out 5 from power and place in front of logarithm,so we would have 5/2*$log_2(2)$,this one

$log_2(2)=1$ so we have left $5/2$,i hope it would help