How do you evaluate #((log_3 (1/27)) - (log_16 4))/((log_5 9)(log_5 125))#?

1 Answer

Answer:

#(log_3 (1/27)-log_16 4)/((log_5 9)(log_5 125))=-log_10 78125/log_10 531441 #

#=-0.85456788#

Explanation:

from the given

#(log_3 (1/27)-log_16 4)/((log_5 9)(log_5 125))#

Let us do this in detail

#log_3 (1/27)=log_3 (1/3^3)=log_3 (3^-3)=-3#

#log_16 4=log_16 4^1=log_16 4^(2/2)=log_16 (4^2)^(1/2) =log_16 (16)^(1/2)=1/2#

#log_5 125=log_5 5^3=3#

Now let us go back to the given

#(log_3 (1/27)-log_16 4)/((log_5 9)(log_5 125))=(-3-1/2)/((log_5 9)*3)=(-7/2)/(3*log_5 9)=(-7/2)/(log_5 729)=(-7/2)/(log_10 729/log_10 5)=(-7/2)*(log_10 5/log_10 729)=-log_10 78125/log_10 531441 #

#=-0.85456788#

God bless .....I hope the explanation is useful.