For log(4)(3x + 4)^5 = 15, do I have to expand (3x + 4)^5, or is there a faster way?
remember that log x^n = n*log x
So,
log(4)(3x + 4)^5 = 15
5log(4)(3x + 4) = 15
log(4)(3x + 4) = 3
3x+4 = 4^3 = 64
3x = 60
x = 20
Expanding (3x+4)^5 would not help, because then you just have a polynomial, but the log of a sum cannot be simplified.
log(x^2+3) is not log x^2 + log 3
That is log 3x^2.