Amateur log problem, forgot how to do!

**Steelers72** · Mar 26th 2013, 04:20 PM

I'm taking a course that requires the use of logs. I forgot how to solve a problem for x like this:

5.75= 4.75 + log [((490(.0230)) +5.90x)/((490(.0215))-5.90x)]

How would you solve this in terms of x (please explain as if teaching a 5 yr old haha..)

Thanks for any help!

**Prove It** · Mar 26th 2013, 05:17 PM

I'll assume that your logarithm has a base of e. Please advise if this is incorrect.

$\displaystyle \begin{align*} 5.75 &= 4.75 + \log{\left( \frac{490 \cdot 0.0230 + 5.9\,x}{490 \cdot 0.0215 - 5.9\,x } \right) } \\ 1 &= \log{ \left( \frac{490 \cdot 0.0230 + 5.9\,x}{490 \cdot 0.0215 - 5.9\,x} \right) } \\ e^1 &= \frac{490 \cdot 0.0230 + 5.9\,x}{490 \cdot 0.0215 - 5.9\,x} \end{align*}$

Go from here.

Thread: Amateur log problem, forgot how to do!

LinkBack

Thread Tools

Display

Amateur log problem, forgot how to do!

Re: Amateur log problem, forgot how to do!

Similar Math Help Forum Discussions

taking trig class and forgot how to do this problem.

Forgot to use ^, here is the Q again. Thank you

Forgot

some how forgot

Fermat's Last Theorem: an amateur proof

Search Tags