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So this is the question:

$\color{darkblue}{h(x)=4\exp(x-4)}\qquad\qquad h^{-1}(x)=\ln\left(\dfrac{\boxed{\phantom{X}}}{\boxed{\phantom{X}}}\right)\,\boxed{\phantom{XXX}}$

It wants me to enter in the inverse function of the log on the left side of the photo. I have my answer, $(e^y + 16)/4$, first of all, I have no idea if it is right or not, secondly, How would I transform it to fit in the areas given in the field provided on the right side? And if my answer is wrong, please do tell my where my mistakes are, there is one fraction, and one box that can be anything, as in $+x$ or $-y$, thank you in advance

Sam Chahine
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  • Your answer is wrong, but how can we tell you where your mistakes are if you haven't told us how you found that answer? – Casteels Dec 04 '13 at 09:55
  • Can you please answer with the correct answer and show working out please, I have been on this question for around two hours now and it's killing me.. I came back from the library hoping to finish it once I got home, but no luck – Sam Chahine Dec 04 '13 at 09:57

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\begin{align*} h(h^{-1}(x)) &= 4e^{h^{-1}(x)-4}\\ x &= 4e^{h^{-1}(x) - 4}\\ \frac{x}{4} &= e^{h^{-1}(x) - 4}\\ \operatorname{ln}\left(\frac{x}{4}\right) &= h^{-1}(x) - 4\\ h^{-1}(x) &=\operatorname{ln}\left(\frac{x}{4}\right) + 4 \end{align*}

JessicaK
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  • Thank you so much. May I ask where would be a good resource to learn this way to complete a question such as this? – Sam Chahine Dec 04 '13 at 09:58
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    The only advice I can give you is to practice. I am hesitant to point you to a source because I don't think it can possibly be sufficiently different from whatever you are already using. – JessicaK Dec 04 '13 at 10:06
  • Thank you, I'l do as many questions of this type until I perfect it, I'l look at your working out for a hint, thanks for the help! – Sam Chahine Dec 04 '13 at 10:08