Prove that log 16 24*4*log 3 2 = 1 + 3*log 3 2

Supposing that 16 and 3 are the bases of logarithms, we'll
note

log3(2)=a.

We'll
transform the base 16 into the base
3:

log16(24)=log3(24)/log3(16)

log16(24)=log3(2^3*3)/log3(2^4)

We'll
use the product rule of
logarithms:

log16(24)=[log3(2^3)+log3(3)]/4log3(2), where
log 3 (3) = 1

log16(24)=[3log3(2)+1]/4log3(2)
(*)

Let's recall that we've
noted log3(2)=a.

We'll substitute log3(2)=a in expression
(*).

(3a+1)/4a=(3a+1)/4a

The
expression from the left side is identical with the expression from the right
side.