Thread: Amateur log problem, forgot how to do!

1. Amateur log problem, forgot how to do!

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!

2. Re: Amateur log problem, forgot how to do!

\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*}