Question

Find the inverse ln(sqrt(x-4))+2 PLEASE HELP ME!

Answer 1

So, we want it in the form x = something, right?

First step, then, is to restate it, including the y:

\[y=\ln \sqrt {x-4}+2\]

We want to rearrange this so that we have x by itself.

The first can be just subtract 2 from each side. This gets us a step closer to having x by itself on one side of the equal sign.

The next step will be to do something to get rid of the ln function. What's the inverse of ln?

Answer 2

For the second step, I would write ln(sqrt(x-4)) as ln(x-4)^1/2 = 1/2ln(x-4).
Then multiply both sides by 2 and then proceed to the next step.

Answer 3

Good simplification.

Answer 4

You still with us? :)

Answer 5

Yeah, that makes sense to me :)

Answer 6

Thanks! :)

Answer 7

what happens next though?

Answer 8

You solve for x in terms of y.
Then interchange x and y. Replace x with y and y with x.

Answer 9

Okay. so would it be 1/2ln(y-4)? What happened to the +2?

Answer 10

\[y = \ln \sqrt {x-4}+2 \\y - 2 = \ln \sqrt {x-4} = \ln(x-4)^{1/2} = \frac 12 * \ln(x-4) \\2(y-2) = \ln(x-4)\]

Answer 11

Why is it 2(y-2)?

Answer 12

\[y - 2 = \frac 12 * \ln(x-4)\]Multiply both sides by 2.\[2(y - 2) = \ln(x-4)\]

Answer 13

Oh okay!

Answer 14

You need to solve for x. Use this identity:
\[\Large If~~\ln(A) = B~~~then~~~A = e^B\]

Answer 15

\[\ln(x-4) = 2(y-2)\\x-4 = e^{2(y-2)}\\x=e^{2(y-2)}+4\\\text{Interchange x and y:}\\y = e^{2(x-2)}+4\]

Answer 16

could it be \[e^{2x-4}\] ??

Answer 17

\[e^2(x-2)\text{ is the same as }e^{2x-4}\]

Answer 18

Also, do you know what the domain and range would be for the inverse we found?

Answer 19

Thank you so much! :)

Answer 20

There is no restriction on the x value in y = e^(2x-4) + 4 and therefore the domain is: (-infinity, infinity).
For the range, the lowest value e^(2x-4) can take is 0 and the highest value is +infinity.
Therefore, the range of e^(2x-4) + 4 is 4 to infinity or [4, infinity)

Answer 21

You are welcome.