Question 1082445
<font face="Times New Roman" size="+2">


The difference of the logs is the log of the quotient, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \ln(x\ +\ 8)\ -\ \ln(5)\ =\ \ln\left(\frac{x\ +\ 8}{5}\right)]


Then, since *[tex \Large \log_b(x_1)\ =\ \log_b(x_2)] if and only if *[tex \Large x_1\ =\ x_2] and *[tex \Large x_i\ \in\ \text{dom} \{\log_b\}]


So set *[tex \Large x\ -\ 4\ =\ \frac{x\ +\ 8}{5}]


and solve for *[tex \Large x] such that *[tex \Large x\ -\ 4\ >\ 0]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>