Use log base 4 on both equations 2 x log4(5) = 3 log4(3) 3 y log4(3) = 2 3 log4(3) = 2 / y 2 x log4(5) = 2 / y log4(5) = 1 / (xy) 4^(log4(5)) = 4^(1/(xy)) 4^(1/(xy)) = 5
From 2nd equation: 4^2 = 3^(3y) -> 4^(2/y} = 3^3 -> 4^(1/y) = 3^(3/2) . From 1st equation: 3^3 = 5^(2x) -> 3^(3/2) = 5^x. Substituting expression for 3^(3/2) from previous line into the starting line gives: 4^(1/y) = 5^x, therefore 4^(1/xy) = 5. No taking logarithms was needed!
Even though Second Method is Simple , we must understand First Method as it gives Comprehensive understanding of log and exponential functions correctly.
I did it in almost a minute by just expanding the given equations to get 25^x=27 and 27^y=16. Using the first equation, raise both sides to the power of y to get 25^xy=27^y=16, so we now have that 25^xy=16. You can check that neither x=0 nor y=0 are solutions, so the number 1/xy is perfectly well defined. Thus, you can raise both sides of the equation to the power of 1/xy to get that 25=16^1/xy. From here we have that 25=4^2/xy, so just square root both sides and you'll get that 5=4^1/xy.
