April 1, 2015

Homework Help: Calculus

Posted by Raj on Sunday, May 6, 2007 at 4:26pm.

Find the inverse of each relation:

y = (0.5)^(x+2)
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.

the answer is y = (log0.5x) - 2

be careful where the bracket is.

for the second, after you interchange the x and y variables you would have
x = 3log2(y-3) + 2
x-2 = log2(y-3)^3

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

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