# Logarithmic question-

1. Aug 3, 2013

### gl0ck

1. The problem statement, all variables and given/known data
Hi, it maybe stupid question, but I struggle on this problem..
log16(4354)+3/log25(43)=a+b*log25

2. Relevant equations

Using properties:
logabn=n logab
logn(bc) = lognb+lognc
logba=1/logab

3. The attempt at a solution
I can't find the b number.
log16(4354)+3/log25(43)=6 log162+4log165+3/3log254=6(log22/log216)+4(log25/log216)+1/log254=3/2 + 2log252= 3/2 + 4log25

so a=3/2 b=4 but in the answer they find b to be 2 not 4
Can you please tell me where is my mistake?

Thanks

Last edited: Aug 3, 2013
2. Aug 3, 2013

### Staff: Mentor

You're missing a right parenthesis, so it's hard to tell what you're starting with.

Did you mean log16(4354) +3/log25(43)?

3. Aug 3, 2013

### gl0ck

Lol, yes, Sorry it is definitely log16(4354)+3/log25(43)

4. Aug 3, 2013

### HallsofIvy

Staff Emeritus
Can we at least presume you know the basic properties of logarithms and so know that $log_{16}(4^35^4)= 3log_{16}(4)+ 4 log_{16}(5)$? And, of course, since $16= 4^2$, $4= 16^{1/2}$ so that $log_{16}(4)= 1/2$. That is, $$log_{16}(4^35^4)= 3+ 4log_{16}(5)$$
$log_{25}(4^3)= 3log_{25}(4)$.

Perhaps what you are missing is that $\log_a(x)= \frac{log_b(x)}{log_a(b)}$. You can use that to change the $log_{25}$ to $log_{16}$. Further, since, $16= 2^4$, $log_{2}(16)= 4$ and so $log_{16}(x)= \frac{log_2(x)}{4}$

You can use that to change everything to $log_2$ as you have on the right.

Last edited by a moderator: Aug 3, 2013
5. Aug 13, 2013

### besulzbach

I'm noob, but I think that you meant
$\log_a(x)= \frac{log_b(x)}{log_b(a)}$
Mainly because if you start with one base and end with two it was not a change of base.
I may be wrong.

6. Aug 13, 2013

### micromass

Staff Emeritus
Don't worry, you're correct :tongue: