How do I solve #(log_3 64) *(log_2 (1/27))# ?

1 Answer
Oct 14, 2017

Answer:

#-18#

Explanation:

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.#