Question 30196
Solve for y: ln(y) + ln(2) = ln(ln(e2)).
ln(y) + ln(2) = ln(ln(e^2)).
ln(y) + ln(2) = ln(2ln(e)).  (using log base e(m^n) = n logbase e (m) )
ln(y) + ln(2) = ln(2).  (since log base e (e) =1 and 2log(e) = 2X1 = 2 )
log (y) = log(2)-log(2) subtracting log(2) from both the sides
log(y) =0
That is y = (e)^0   (applying definition log base e (N) = p implies N= (e)^p)
That is y = 1
Answer: y = 1
Verification: can be done orally(using log(1) =0 )