Question

PLEASE PLEASE PLEASE PLEASE HELP!!!
How can I solve an exponetial model to make an inverse one ?

Answer 1

That would depend on the exponential model. Please provide one and give it your best effort.

Answer 2

h(t)=106(0.95)^t+50

Answer 3

I got as far as \[t-50\div160=0.95^{h}\]

Answer 4

Pretty good, excepting that you have forgotten your Order of Operations.

I hope you mean \((t-50)/160 = 0.95^{h}\). Without the parentheses, it is quite a different matter.

Have you considered introducing a logarithm?

Answer 5

Yes, which is where I get confused

Answer 6

log\[\log_{0.95}(t-50)\div160=\log_{0.95}(0.95)^h\]

Answer 7

Logs base 0.95?!! Why on earth would you do that? Use somethin normal.

\(\log\left(\dfrac{t-50}{160}\right) = \log\left(0.95^{h}\right)\)

Now, use logarithm rules.

\(\log\left(\dfrac{t-50}{160}\right) = h\cdot\log\left(0.95\right)\)

You're almost done.

Note: You can use the logs base 0.95 if you want, but I'm guessing that will be more confusing.

Answer 8

Okay so for the the side with H I would get -0.022

Answer 9

How do I get rid of that ?

Answer 10

?

Answer 11

Why did you get that at all? \(\log(0.95)\) is just fine.

Answer 12

and that would be my inverse function ?

Answer 13

Well, did you solve for \(h\)?

Answer 14

h=1/160 log(t-50)/log(.95)

Answer 15

Is that simplified enough ?

Answer 16

?? How did the 160 get outside the logarithm? If you were to use logarithm rules, you should get \(\log((t-50)/160) = \log(t-50) - log(160)\).

Answer 17

You can also use the base transformation to rewrite a little, but it's still not very pretty.

\(\dfrac{\log\left(\dfrac{t-50}{160}\right)}{\log(0.95)} = \log_{0.95}\left(\dfrac{t-50}{160}\right) = h\)

Answer 18

okay thank you