I need help in solving log question? Please help.

If logx = 0.6 and logy = 0.2, evaluate log #(x^2/y)#

1 Answer
Feb 23, 2018

We need to find the value of, #log(x^2/y)#

#=logx^2 - logy#

#=2logx - logy#

Note: #color(red)(log(a/b) = loga-logb# and #color(red)(logx^y = ylogx# are applied.

Now, plug in the given values.

#=2(0.6) - (0.4)#

#=1.2-0.4 = 0.8#