We should see that the bases are mismatched:
64=2^(2*3)
1/27=3^-3
So, we should apply this property of logs to change the bases:
log_ax=(log_b(x))/(log_b(a))
Where a is the original base and b is the new base.
We need to change log_3x to log_2x and vice versa. So:
log_3(64)=(log_2(64))/(log_2(3))
log_2(1/27)=(log_3(1/27))/(log_3(2))
log_3(64)*log_2(1/27)=(log_2(64))/(log_2(3))*(log_3(1/27))/(log_3(2))
Since log_a(b)*log_b(a)=1
=(6*-3)/1
=-18
Proving some things:
Let's prove that log_ax=(log_b(x))/(log_b(a))
Let's start with a^y=x
Take log_b of both sides, where b is the target base:
log_b(a^y)=log_b(x)
ylog_b(a)=log_b(x)
y=(log_b(x))/(log_b(a))
We know that in a^y=x, y=log_a(x) therefore:
log_a(x)=(log_b(x))/(log_b(a))
Q.E.D.
Let's prove that log_a(b)*log_b(a)=1
Let a^x=b, b^y=a prove xy=1 so y=1/x
Therefore with a^x=b prove b^(1/x)=a
Rewrite with logs:
log_a(b)=x and prove log_b(a)=1/x
change base from b to a:
log_b(a)=(log_a(a))/(log_a(b))=1/(log_a(b))
Since we know that log_a(b)=x
log_b(a)=1/x
Q.E.D.
