Log_e (5) = 1.609 Explanation: Log_e (5) = ln(5) = 1.609 where n stands for natural logarithm. The natural logarithm is such that ln(e) =1. Alternatively, Log_e(5) is the number such that e^x = 5.

This is a constant. The derivative of any constant, with respect to any variable, is 0. Explanation: \displaystyle{{\log}_{{a}}{a}}={1} , \displaystyle{{\log}_{{e}}{e}}={\left({{\log}_{{2}}{e}}\right)}{\left({{\log}_{{e}}{2}}\right)} ...

The alternating series theorem gives you a bound. For a series where the signs alternate and the terms decrease in magnitude the error is always of the sign of the first neglected term and smaller ...

Both of your answers are correct as I pointed out in my comment:\log_3 e=\frac{\log_e e}{\log_e 3}=\frac{1}{\log_e 3}

Could make it simpler: \begin{align*} \ln (a+bz) &= \ln \left[ a\left(1+\frac{bz}{a} \right) \right] \\ &= \ln a+\ln \left(1+\frac{bz}{a} \right) \\ &= \ln a+\sum_{n=1}^{\infty} ...

You can always write x=b^{\log_bx} when x,b>0. So that means that if you felt like it you could indeed write x=b^{\log_bx}=b^{\log_bb^{\log_bx}}=b^{\log_bb^{\log_bb^{\log_bx}}}=\ldots ...