Solving $\frac{3.75^{a+1}}{5^{a+1}}=0.4871 $ Having a brain fart, and don't know how to solve for the variable here. What can I do to simplify this equation to figure out a.
$$\frac{3.75^{a+1}}{5^{a+1}}=0.4871 $$
Just a little bit confused as to how to isolate a here.
Attempt:
$\frac{3.75^{a+1}}{5^{a+1}}=0.4871$
$ \frac{3.75^a\times3.75}{5^a\times 5} = 0.4871$
$ \frac{3.75^a\times3.75}{5^a\times 5} = 0.4871$
$\frac{3.75^a}{5^a} = 0.649466667$
When it's not base e, I'm sure I use log but not entirely sure from here where to go.
 A: We have
\begin{align}
0.4871 &= \dfrac{3.75^{a+1}}{5^{a+1}} \\
&= \left( \dfrac{3.75}{5} \right)^{a+1} \\
&= \left(\dfrac{3}{4} \right)^{a+1}
\end{align}
Hence, taking $\ln$ of both sides (base $e$), and using the fact that $\ln(\alpha^\beta) = \beta \ln(\alpha)$, we get
\begin{align}
(a+1) \ln \left(\dfrac{3}{4} \right) = \ln(0.4871)
\end{align}
Hence,
\begin{align}
a &= \dfrac{\ln(0.4871)}{\ln(3/4)} -1
\end{align}

Of course, there is no particular reason to take logarithm to base $e$ (any other base is fine), but that's just something mathematicians like, as opposed to some other base.

Alternatively, we can continue from where you left off:
\begin{align}
\dfrac{3.75^a}{5^a} = 0.649...
\end{align}
If for some reason you forget that you can group the powers like I did above, you can still take $\ln$ of both sides. But now, you need to remember that $\ln\left(\frac{\alpha}{\beta} \right) = \ln(\alpha) -\ln(\beta)$. Hence,
\begin{align}
\ln(0.649...) &= \ln(3.75^a) - \ln(5^a) \\
&= a \ln(3.75) - a \ln(5) \\
&= a \left[ \ln(3.75) - \ln(5)\right] 
\end{align}
Hence, you can divide to get
\begin{align}
a &= \dfrac{\ln(0.649...)}{\ln(3.75) - \ln(5)}
\end{align}
You can re combine the denominator by reversing the logarithm rules to get $\ln(3/4)$ in the denominator, to make it look more like my answer above. But if you're looking for an exact answer, then of course, you should express the $0.649...$ as an exact fraction.
A: $$\frac{3.75}5=0.75$$
So the equation is equivalent to
$$(0.75)^{a+1}=0.4871$$
Taking logs both sides gives
$$(a+1)\ln{(0.75)}=\ln{(0.4871)}$$
$$\therefore a=\frac{\ln{(0.4871)}}{\ln{(0.75)}}-1\approx1.500280368$$
