1. ## Question about e^2x graph

Use a graph produced by your calculator to estima the values of $\displaystyle x$ for which $\displaystyle e^2^x$ is smaller than $\displaystyle (x-8)^4$. Also, describe clearly and accurately the purpose of each needed step.

This is what I tried but I am totally lost because I can't figure out how to approach this:
$\displaystyle e^2^x<(x-8)^4$

I could take the 4th root of both sides but I can't isolate $\displaystyle x$. Not too sure how to solve this question.

2. ## Re: Question about e^2x graph

But beside the excercice you need the graphs of the functions, so have you tried to solve this exercice with the calculator? ...

3. ## Re: Question about e^2x graph

Try $\displaystyle e^{2x}- (x-8)^4$ where are the zeros?

4. ## Re: Question about e^2x graph

Originally Posted by freestar
Use a graph produced by your calculator to estima the values of $\displaystyle x$ for which $\displaystyle e^2^x$ is smaller than $\displaystyle (x-8)^4$. Also, describe clearly and accurately the purpose of each needed step.

This is what I tried but I am totally lost because I can't figure out how to approach this:
$\displaystyle e^2^x<(x-8)^4$

I could take the 4th root of both sides but I can't isolate $\displaystyle x$. Not too sure how to solve this question.
It says use a calculator for that reason - plot both graphs on the same axes and pick the appropriate values.

You know that $\displaystyle (x-8)^4 \geq 0$ and that the points (0,8^4) and (8,0) lie on it.
You know that $\displaystyle e^{2x}$ has the x-axis as an asymptote and the point (0,1) lies on it.

Here is the wolfram plot, I have limited the domain to make it easier to see: plot &#123;y&#61;e&#94;&#40;2x&#41;,y&#61;&#40;x-8&#41;&#94;4&#125; between 0 and 5 - Wolfram|Alpha

5. ## Re: Question about e^2x graph

Originally Posted by e^(i*pi)
It says use a calculator for that reason - plot both graphs on the same axes and pick the appropriate values.

You know that $\displaystyle (x-8)^4 \geq 0$ and that the points (0,8^4) and (8,0) lie on it.
You know that $\displaystyle e^{2x}$ has the x-axis as an asymptote and the point (0,1) lies on it.

Here is the wolfram plot, I have limited the domain to make it easier to see: http://www.wolframalpha.com/input/?i=plot+{y%3De^%282x%29%2Cy%3D%28x-8%29^4}+between+0+and+5
So in that case, $\displaystyle x\leq3.155633123$?

Correct!

7. ## Re: Question about e^2x graph

Originally Posted by freestar
So in that case, $\displaystyle x\leq3.155633123$?
I'd say so, it agrees with the general value of the plot and I assume your calculator spat it out