1. ## Logarithms

How would I go about solving this problem?

$\displaystyle 2 = e^5x$ (5x being the exponent)

The whole natural logarithms, antilogarithms, and Euler's number stuff isn't really clicking in my head.
If you explained it step by step that would help a lot.

2. Originally Posted by Hypertension
How would I go about solving this problem?

$\displaystyle 2 = e^5x$ (5x being the exponent)

The whole natural logarithms, antilogarithms, and Euler's number stuff isn't really clicking in my head.
If you explained it step by step that would help a lot.
you mean $\displaystyle 2 = e^{5x}$..

so here are some reviews..

$\displaystyle \ln e = 1$

$\displaystyle \ln a^{b} = b \ln a$

this must be sufficient..

first, take the natural logarithms of both sides..

3. so,

$\displaystyle 2 = e^{5x}$

=> $\displaystyle (ln)2 = (ln)e^{5x}$

=> $\displaystyle .6931 = 5x ln e$

=> $\displaystyle .6931 = 5x$ (since $\displaystyle ln e = 1$, right?)

=> $\displaystyle \frac{.6931}{5} = \frac{5x}{5}$

=> $\displaystyle x = .1386$

?

4. bravo!!

anyways, put \ before ln i.e
Code:
\ln a
will look like $\displaystyle \ln a$

5. ok. And thanks