Given that f(x) is 4^8_+1 what is f^-1_(x)?
- az_lenderLv 78 months agoFavourite answer
If you meant
f(x) = 4^(8x+1), then
8x+1 = log_4(f(x)), and
x = [log_4(f(x)) - 1]/8, so
f^(-1) (x) = [log_4(x) - 1]/8.
- Anonymous8 months ago
f(x) = 4⁸ + 1
f⁻¹(x) = ?f(x) = yx = f⁻¹(y)y = 4⁸ + 1Here there is ‘x’ termLet f(x) = 4x + 1f⁻¹(x) = ?y = 4x + 1y-1 = 4xlog(y-1) = xlog4x = log(y-1) / log4x = log(y-1-4)x = log(y-5)f⁻¹(x) = log(x-5).Source(s): http://myrank.co.in/
- AshLv 78 months ago
There is no x in f(x). Can you please check your problem ?