The problem is:

$\displaystyle e^{lnx} = 4$

Is there a way I can rewrite this to solve for x, or some rule that I am suppose to apply?

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- Apr 30th 2012, 04:28 PMBashyboyExponential equation
The problem is:

$\displaystyle e^{lnx} = 4$

Is there a way I can rewrite this to solve for x, or some rule that I am suppose to apply? - Apr 30th 2012, 05:29 PMskeeterRe: Exponential equation
- Apr 30th 2012, 05:39 PMPlatoRe: Exponential equation
- May 1st 2012, 04:58 AMBashyboyRe: Exponential equation
@Plato:

Well, clearly it is not a simple problem for me, especially seeing that I have not dealt with a problem like this for two semester. And if it is such a waste, why spend the time replying?

@Skeeter:

Now I believe I remember seeing something like this. And isn't true that the base e and the exponent go away because they are inverses? That's the part I have a little trouble wrapping my head around. That's the only explanation I have been given, and it just seem a good enough explanation for me.

What I started to do was: $\displaystyle ln(e)^{lnx} = ln(4)\rightarrow\ln(x)\times\ln(e) = ln(4)\rightarrow\ln(x) = ln(4)$, then finally $\displaystyle e^{ln(4)} = x$

which sort of brought me back to square one. I know I can just evaluate it in the calculator, but I don't know why the answer comes out the way it does. - May 1st 2012, 08:15 AMskeeterRe: Exponential equation
$\displaystyle y = \ln{x}$

written as an equivalent exponential function ...

$\displaystyle x = e^y$

... but what is y in the first place? - May 1st 2012, 10:37 AMHallsofIvyRe: Exponential equation
BashyBoy, I think that Plato was a bit annoyed that you would post a question like that, involving exponential and logarithms, without finding out, or reviewing, what exponentials and logarithms

**are**. And the whole point of the problem is the**definition**of 'exponential' and 'logarithm'.