# Calculus

posted by
**Raj**
.

Find the inverse of each relation:

y = (0.5)^(x+2)

and

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

For the first one I got y=log base 0.5 (x+2)...but the answer in the back of the textbook says that it is not x+2, but x-2. Can someone tell me why it would end up being x-2 and help me with the second one too.

I agree with your first answer.

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

y-2 = log3 (x-3)^3

take the log3 of each side..

log3 (y-2)= (x-3)^3

take the cube root of each side

1/3 log3 ((y-2)) = x-3

x= 1/3 log3 ((y-2))+3 but 3 is log3 (27)

x= 1/3 log3( y-2) + log2 (27)

= 1/3 log3 (27*(y-3))

check that.

y=

the answer is y = (log_{0.5}x) - 2

be careful where the bracket is.

for the second, after you interchange the x and y variables you would have

x = 3log_{2}(y-3) + 2

x-2 = log_{2}(y-3)^3

(y-3)^3 = 2^{y-3}

y+3 = [2^{y-3}]^(1/3)

y = [2^{y-3}]^(1/3) - 3

the exponent on 2 inside the square bracket in the last 3 lines should have been x-2 instead of y-3

I kept copy-and-pasting so I kept copying my own typo.

If y is the x+2 power of 0.5, then you are correct. By definition, y is the log-to-base 0.5 of x+2

But they are asking for the inverse. What function of y is x?

(x+2) log 0.5 = log y (to any base)

x = [log y/log(0.5)] -2 (to any base)

= log(base0.5)y - 2

Yea, I figured out the first one.

y = (0.5)^(x+2)

x = (0.5)^(y+2)

log(base 0.5)x = y+2

log(base 0.5)x-2 = y

For the second one, Reiny, you're right.

Thanks all for the help.