How to resolve the logarithm?

Question

I have this exercise:

My result is : $16/3$, but it does not appear in the answers, what is wrong?

Or is it another representation of the same logarithm?

score 0 · Accepted Answer · answered Jan 16 '18 at 14:51

0

Let apply the usual rules for logarithm

$$\log_a b=\frac{1}{\log_b a} \quad \quad \log_a b^c=c\cdot \log_a b$$

thus

$$\log_{27}3+\log_2 32=\frac{1}{\log_3 3^3}+\log_2 2^5=\frac13+5=\frac{16}3$$

answered Jan 16 '18 at 14:51

user

So the guide is wrong ? – Eduardo Campos Jan 16 '18 at 15:40
@EduardoCampos It seems so! By inspection I can't recognize any other answer s correct! – user Jan 16 '18 at 15:43
Thanks for explanation – Eduardo Campos Jan 16 '18 at 16:15
You are welcome ! – user Jan 16 '18 at 16:28

score 0 · Answer 2 · answered Jan 16 '18 at 14:51

0

I think you are right: $$\log_{27}3+\log_232=\frac{1}{3}+5.$$

answered Jan 16 '18 at 14:51

Michael Rozenberg

So the guide is wrong ? – Eduardo Campos Jan 16 '18 at 15:40
I think the answer in the book is wrong. – Michael Rozenberg Jan 16 '18 at 15:42

2 Answers2