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 = (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.
To find the inverse of a relation, you need to switch the roles of x and y and solve for y.
For the first relation:
You correctly found that the inverse is y = log base 0.5 (x + 2). However, the textbook states that the answer is y = log base 0.5 (x - 2). Let's see why.
Starting with the original equation, y = (0.5)^(x + 2), we switch x and y:
x = (0.5)^(y + 2).
To solve for y, we need to isolate it. Taking the logarithm of both sides with base 0.5:
log base 0.5 (x) = y + 2.
Now, subtract 2 from both sides:
log base 0.5 (x) - 2 = y.
Thus, the inverse relation is y = log base 0.5 (x - 2). This is the same as the answer in the textbook.
For the second relation:
The given equation is y = 3log base 2 (x - 3) + 2. To find the inverse, we switch x and y:
x = 3log base 2 (y - 3) + 2.
To solve for y, we follow a similar process. First, subtract 2 from both sides:
x - 2 = 3log base 2 (y - 3).
Next, divide both sides by 3:
(x - 2)/3 = log base 2 (y - 3).
To eliminate the logarithm, we can rewrite the equation using exponentiation:
2^((x - 2)/3) = y - 3.
Finally, add 3 to both sides to isolate y:
2^((x - 2)/3) + 3 = y.
So, the inverse relation is y = 2^((x - 2)/3) + 3.