What is the inverse of h(x)=12e^2x? Based on our answer, write an expression equivalent to x where h(x)=10
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
\[a) \frac{ x }{ 12 }=e^(2y)\]
Im not sure how to do B
yup
oh ok
\[e^A=B\iff A=\ln(B)\]
replace \(A\) by \(2y\) and \(B\) by \(\frac{x}{12}\)
\[2y=\ln \frac{ x }{ 12 }\] Is tha correct?
yes
although i would write \[2y=\ln(\frac{x}{12})\] since the log is a function
Oh. Then it's: \[y= \ln \frac{ x }{ 6 }\] ?
oh no!!
hold on, lets go slow
For Part C, divide by 2, am I wrong? ;o
you see how you wrote \[2y=\ln \frac{ x }{ 12} \] and i said "i would write \[2y=\ln \left(\frac{ x }{ 12}\right) \]
Oh yes. I thought that didn't make any difference :#. I guess now I see the value in parenthesis
it is , in english, the log OF (x over 12) not "the log of x" over 12
like for example if \(x=3\) then it would be \[\log(\frac{3}{12})=\log(\frac{1}{4})\]
half of \[\ln(\frac{x}{12})\] is \[\frac{1}{2}\ln(\frac{x}{12})\]
Oh damn. I made a critical error
or even \[\frac{\ln(\frac{x}{12})}{2}\]
clear right? just like if \[f(x)=x^2\] then \[\frac{1}{2}f(x)=\frac{x^2}{2}\] not \[\left(\frac{x}{2}\right)^2\]
yes, very clear
ok good, then we are done
Wait. If we do the second part of the question. We do \[10=\frac{ 1 }{ 2 } \ln (\frac{ x }{ 12 })\]?
no
\[h(x)=12e^{2x}\] so \[h(x)=10\] means \[12e^{2x}=10\]
\\[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
is plain englsh if \(h(x)=10\) then\[x=\frac{1}{2}\ln(\frac{10}{12})\]
OH. And now we just plug it in the calculator, right?
WAIT. IT SAYS EXPRESSION. Thanks for clarifiyng!
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