math help please?

Question

how do find f^-1 (inverse) for the function f?

f(x) = logˇ2 (-x+3)-5
{log-base-2 (-x+3)-5 }

f(x) = e^(x+2) -3

please show how you got the answer
thank you

Answer 1

1) 2^(y+5)=(-x+3)
so x=3-2^(y+5) interchange x and y so
y= 3-2^(x+5) ( I supposed -5 is not included in log
2)ln (y+3) = x+2 so
x= ln(y+3) -2 and y = ln(x+3)-2

Answer 2

f(x) = y = log(2)(-x+3)-5

log(2)(3-x) = y + 5

2^(y+5) = 3-x

x = 3 - 2^(y+5)

f^-1(x) = 3 - 2^(x+5)

2)

f(x) = y = e^(x+2) - 3

e^(x+2) = y + 3

x+ 2 = log(y+3)

x = log(y+3) - 2

f^-1(x) = log(x+3) - 2

Answer 3

i will think abt it