evaluate log expression

Question

log x^(6/5)*y^(4/5)

given that log x=2, log y =3. log 2 =0.3, and log 3=0.48

the answer is 24/5 but I dont know what to do after getting rid of the radical

Answer 1

Hi Doris,

 

Based on the product identity, we can write

log (x^(6/5)*y^(4/5)) = log x^(6/5) + log y^(4/5)

Then, based on power identity, bring the powers to the front

log x^(6/5) + log y^(4/5) = 6/5*log x + 4/5*log y

you are given log x and log y so just plug them in

6/5*log x + 4/5*log y = (6/5)*2 + (4/5)*3

= 12/5 + 12/5

= 24/5