Using logs to solve an inequality

Question

When I use a natural log to solve $5000 (0.88)^{(n-1)}$ < $100$ I get the correct answer $n > 31.6$ so $n = 32$ But when I use log to the Base $0.88$ I get $n < 31.6$ The inequality sign is incorrect, why?

Answer 1

Because $\log_{0.88}$ is not what you think it is (in short: it is a decreasing function, not increasing as $\log_2,\log_{10}$ are). The base $b$ of a logarithm must be a positive number different than $1$, and then we have

$$ \log_b x = \frac{\ln x}{\ln b} $$ where $\ln$ denotes the natural logarithm (base $e$). (Cf. e.g. the Wikipedia page.)

Now, observe that for $b=0.88$, we have $\ln b < 0$ (since $b\in(0,1)$). So, when you get (solving the inequality) $$ (n-1) < \log_{0.88}\frac{100}{5000} $$ you should see that you applied the function $\log_{0.88}=\frac{\ln}{\ln 0.88}$ to both sides of the original inequality, and this function $\log_{0.88}$ is a decreasing function (since $\ln$ is increasing). You should have switched the sign of the inequality.

(Indeed, if $f$ is a decreasing function, then $a< b$ is equivalent to $f(a)>f(b)$: the sign only remains the same for increasing functions.)

Answer 2

$\log_a x = \frac {\ln x}{\ln a}$ and if $a<1, \ln a < 0$

When you take the log in base $0.88$ you are effectively dividing by a negative number, and that flips the direction of your inequalities.