Logarithmic Question: Solving for Unknowns Using Logarithm Properties"

[gl0ck]

Homework Statement

Hi, it maybe stupid question, but I struggle on this problem..
log₁₆(4³5⁴)+3/log₂₅(4³)=a+b*log₂5

Homework Equations

Using properties:
log_abⁿ=n log_ab
log_n(bc) = log_nb+log_nc
log_ba=1/log_ab

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

 

[Mark44]

Homework Statement

Hi, it maybe stupid question, but I struggle on this problem..
log₁₆(4³5⁴+3/log₂₅(4³)=a+b*log₂5

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)?

Homework Equations

Using properties:
log_abⁿ=n log_ab
log_n(bc) = log_nb+log_nc
log_ba=1/log_ab

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

 

[gl0ck]

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

 

[HallsofIvy]

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.

 

[besulzbach]

Perhaps what you are missing is that $\log_a(x)= \frac{log_b(x)}{log_a(b)}$

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.

 

[micromass]

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.

Don't worry, you're correct