Log problem: evaluate log-base-16(32)

[Guest]

What is the answer to this question?

log 16 (as a subscript) 32 = ?

My options are 5/2, -5/2, 5, -5, 2 and none of the above. I chose none of the above and I missed it.

 

[royhaas]

$\displaystyle log_{16}(32)=log_{16}(16)+log_{16}(2)=1 +(1/4)log_{16}(2^4)$.

 

[Guest]

Log problem

On my log problem,what is the final answer then?

 

[pka]

The correct answer in you list is "none of the above".
Do you see why?

 

[Guest]

I chose none of the above for the test question and the professor said that no that was wrong. So again, my options were 5/2, -5/2, 2 or -2.


Thanks.

 

[pka]

I chose none of the above for the test question and the professor said that no that was wrong. So again, my options were 5/2, -5/2, 2 or -2.

Go tell the professor that s/he is the one who is wrong.
I suspect that s/he thinks that the correct answer is 5/2.
But as you can see it is 5/4.
\(\displaystyle \L
\left( {\sqrt[4]{{16}}} \right)^5 = \left( {16} \right)^{\frac{5}{4}} \quad \Rightarrow \quad \log _{16} \left( {32} \right) = \frac{5}{4}\)

 

[soroban]

Hello, biokimmi!

What is the answer to this question?

$\displaystyle \;\;\;\log_{16}(32)\:=\:?$

My options are: $\displaystyle \frac{5}{2},\;-\frac{5}{2},\;5,\:-5,\;2,$ and $\displaystyle \text{none of the above}$.

I chose none of the above and I missed it.

Unless there's a typo, your answer is correct.


We have:$\displaystyle \;\log_{16}(32) \:=\:x$

Rewrite in exponential form: $\displaystyle \,16^x\:=\:32$

Try to get the same base: $\displaystyle \,(2^4)^x\:=\:2^5$

And we have: $\displaystyle \,2^{4x}\:=\:2^5$


We have equal bases, so we can equate the exponents: $\displaystyle \,4x\:=\:5\;\;\Rightarrow\;\;x\,=\,\frac{5}{4}$

Therefore: \(\displaystyle \L\,\log_{16}(32)\:=\:\frac{5}{4}\)

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If you doubt it, use the Base-Change Formula:

\(\displaystyle \L\;\;\;\log_{16}(32)\;=\;\frac{\log(32)}{\log(16)} \;=\;1.25\) . . . see?