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.

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.
\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.
$$\frac{3.75}5=0.75$$ So the equation is equivalent to $$(0.75)^{a+1}=0.4871$$ Taking logs both sides gives $$(a+1)\ln{(0.75)}=\ln{(0.4871)}$$ $$\therefore a=\frac{\ln{(0.4871)}}{\ln{(0.75)}}-1\approx1.500280368$$