Solve completely for a (logrithm with an unknown base)

Question

I'm trying to figure out how to calculate the base if: $$\log_A 270- \frac{\log_A 10}{\log_A \frac12} = 2$$ how can I find $A$? If someone could let me know the steps to find $A$ in plain english, I'd be eternally grateful! thank you

Answer 1

The easiest way I think would be with the change of base formula $\log_x y=\frac{\log_B y}{\log_B x}$ for your favourite choice of base $B$. Using this to write your equation:

\begin{align*} &\log_A270-\frac{\log_A10}{\log_A\frac{1}{2}}=2\\ \implies&\frac{\log_e270}{\log_eA}-\frac{\left(\frac{\log_e10}{\log_eA}\right)}{\left(\frac{\log_e\frac{1}{2}}{\log_eA}\right)}=2\\ \implies&\frac{\ln270}{\ln A}-\frac{\ln10}{\ln\frac{1}{2}}=2\\ \implies&\frac{\ln270}{\ln A}+\frac{\ln10}{\ln2}=2\\ \implies&\frac{\ln270}{\ln A}+\frac{\ln2+\ln5}{\ln2}=2\\ \implies&\frac{\ln270}{\ln A}+\frac{\ln5}{\ln2}=1\\ \implies&\ln A=\frac{\ln270}{1-\frac{\ln5}{\ln2}}\\ \implies&A=e^{\left(\frac{\ln270}{1-\frac{\ln5}{\ln2}}\right)}\\ \implies&A\approx0.014479 \end{align*}