How to solve this simple exponential equation?

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February 25th 2011, 03:23 AM
Yehia

How to solve this simple exponential equation?

(16e^x) - (16e^2x) = 3

and the answer says that x = 0.25 but i can't see how you can get there? please help, much appreciated!! thanks!
February 25th 2011, 03:27 AM
Prove It

Rewrite it as $\displaystyle 0 = 16e^{2x} - 16e^x + 3$

$\displaystyle 0 = 16\left(e^x\right)^2 - 16e^x + 3$ .

Now let $\displaystyle X = e^x$ so that the equation becomes

$\displaystyle 0 = 16X^2 - 16X + 3$ , which is a quadratic equation you can solve for $\displaystyle X$ , and then solve for $\displaystyle x$ .
February 25th 2011, 03:32 AM
FernandoRevilla

Quote:

Originally Posted by Yehia

(16e^x) - (16e^2x) = 3 and the answer says that x = 0.25 but i can't see how you can get there? please help, much appreciated!! thanks!

Have you correctly quoted the problem?
February 25th 2011, 04:27 AM
Yehia

Quote:

Originally Posted by Prove It

Rewrite it as $\displaystyle 0 = 16e^{2x} - 16e^x + 3$

$\displaystyle 0 = 16\left(e^x\right)^2 - 16e^x + 3$ .

Now let $\displaystyle X = e^x$ so that the equation becomes

$\displaystyle 0 = 16X^2 - 16X + 3$ , which is a quadratic equation you can solve for $\displaystyle X$ , and then solve for $\displaystyle x$ .

thanks
February 25th 2011, 05:14 AM
FernandoRevilla

Quote:

Originally Posted by Yehia

(16e^x) - (16e^2x) = 3 and the answer says that x = 0.25 but i can't see how you can get there? please help, much appreciated!! thanks!

Thank you very much for answering to my previous question. Please, post your work if you get some of this

(i) $x=0,25$ is a solution of the given equation.

(ii) The equation has only one solution.

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