Question

I am having trouble figuring out how to solve these logarithms. Could someone please help!

log2(log4x)=1 and

solve for x and y:
(1/2)^x+y= 16 logx-y8=-3

Answer 1

log2(log4x)=1
Rewrite as
(log((log(x))/(log(4))))/(log(2)) = 1
Divide both sides by 1/(log(2)):
log((log(x))/(log(4))) = log(2)
Cancel logarithms by taking exp of both sides:
(log(x))/(log(4)) = 2
Divide both sides by 1/(log(4)):
log(x) = 2 log(4)
Cancel logarithms by taking exp of both sides:
x = 16

You need to rewrite the 2nd with parentheses.

As written it is not clear.
(1/2)^x+y= 16 logx-y8=-3