Question

How do you evaluate 1/3log 8 ?​

Answer 1

Answer:

We can write log16 as log24=4log2 and log8 as log23=3log2. This gives us

logx=12⋅4log2−3⋅13log2+1.

A bit of simplifying yields

2log2−log2+1=log2+1

But note that log10=1 so we get

log2+log10=log(2⋅10)=log20

So logx=log20⟺x=20 by