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Mathematics 21 Online
OpenStudy (goldphenoix):

What is the inverse of h(x)=12e^2x? Based on our answer, write an expression equivalent to x where h(x)=10

satellite73 (satellite73):

solve \[x=12e^{2y}\] for \(y\) in three steps a) divide both sides by 12 b) write in equivalent logarithmic form c) divide both sides by 2

OpenStudy (goldphenoix):

\[a) \frac{ x }{ 12 }=e^(2y)\]

OpenStudy (goldphenoix):

Im not sure how to do B

satellite73 (satellite73):

yup

satellite73 (satellite73):

oh ok

satellite73 (satellite73):

\[e^A=B\iff A=\ln(B)\]

satellite73 (satellite73):

replace \(A\) by \(2y\) and \(B\) by \(\frac{x}{12}\)

OpenStudy (goldphenoix):

\[2y=\ln \frac{ x }{ 12 }\] Is tha correct?

satellite73 (satellite73):

yes

satellite73 (satellite73):

although i would write \[2y=\ln(\frac{x}{12})\] since the log is a function

OpenStudy (goldphenoix):

Oh. Then it's: \[y= \ln \frac{ x }{ 6 }\] ?

satellite73 (satellite73):

oh no!!

satellite73 (satellite73):

hold on, lets go slow

OpenStudy (goldphenoix):

For Part C, divide by 2, am I wrong? ;o

satellite73 (satellite73):

you see how you wrote \[2y=\ln \frac{ x }{ 12} \] and i said "i would write \[2y=\ln \left(\frac{ x }{ 12}\right) \]

OpenStudy (goldphenoix):

Oh yes. I thought that didn't make any difference :#. I guess now I see the value in parenthesis

satellite73 (satellite73):

it is , in english, the log OF (x over 12) not "the log of x" over 12

satellite73 (satellite73):

like for example if \(x=3\) then it would be \[\log(\frac{3}{12})=\log(\frac{1}{4})\]

satellite73 (satellite73):

half of \[\ln(\frac{x}{12})\] is \[\frac{1}{2}\ln(\frac{x}{12})\]

OpenStudy (goldphenoix):

Oh damn. I made a critical error

satellite73 (satellite73):

or even \[\frac{\ln(\frac{x}{12})}{2}\]

satellite73 (satellite73):

clear right? just like if \[f(x)=x^2\] then \[\frac{1}{2}f(x)=\frac{x^2}{2}\] not \[\left(\frac{x}{2}\right)^2\]

OpenStudy (goldphenoix):

yes, very clear

satellite73 (satellite73):

ok good, then we are done

OpenStudy (goldphenoix):

Wait. If we do the second part of the question. We do \[10=\frac{ 1 }{ 2 } \ln (\frac{ x }{ 12 })\]?

satellite73 (satellite73):

no

satellite73 (satellite73):

\[h(x)=12e^{2x}\] so \[h(x)=10\] means \[12e^{2x}=10\]

satellite73 (satellite73):

\\[h^{-1}=\frac{1}{2}\ln(\frac{x}{12})\] so you solve \[h(x)=10\] by taking \[h^{-1}(10)\] that was the point of the problem , now that you have the inverse, you can solve

satellite73 (satellite73):

is plain englsh if \(h(x)=10\) then\[x=\frac{1}{2}\ln(\frac{10}{12})\]

OpenStudy (goldphenoix):

OH. And now we just plug it in the calculator, right?

OpenStudy (goldphenoix):

WAIT. IT SAYS EXPRESSION. Thanks for clarifiyng!

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