How do I solve for $t$ in this indices question $20(10^{0.1t})=25(10^{0.05t})$ How do I solve for $t$ in this indices question?
$$20(10^{0.1t})=25(10^{0.05t})$$
I have tried using the log rules and isolating $t$, but I could not seem to find the answer. Can someone please show some working out for me as I suspect I may be messed on my working process. Thank you
 A: Taking logs of both sides, then splitting the products into sums by logarithm rules, yields
$$\log(20\cdot10^{0.1t})=\log(25\cdot10^{0.05t})$$
$$\log20+0.1t=\log25+0.05t$$
$$0.05t=\log25-\log20=\log\frac54$$
$$t=20\log\frac54=\log\frac{5^{20}}{4^{20}}$$
A: Hint:  $$20\cdot10^{\frac{1}{10}t}=25\cdot10^{\frac{1}{20}t}\iff
20\cdot(10^{\frac{1}{20}t})^2-25\cdot10^{\frac{1}{20}t}=0
\iff5\cdot10^{\frac{1}{20}t}\bigl(4\cdot10^{\frac{1}{20}t}-5\bigr)=0.$$ Now factorize and take logarithm.
ALternatively maybe even simpler: Divide by $10^{\frac{1}{20}t}$ and by $20$ to get
$$10^{\frac{1}{20}t}=\frac54.$$
Now take logarithm.
A: Note that


*

*$10^{0.1 t} = 10^{2 \cdot 0.05 t}$
$$\Rightarrow 20(10^{0.1t})=25(10^{0.05t}) \Leftrightarrow \frac{10^{2 \cdot 0.05 t}}{10^{0.05 t}} = \frac{25}{20} \Leftrightarrow 10^{0.05 t} = \frac{5}{4} \stackrel{\mbox{for} \log_{10}}{=}\frac{10}{8}$$
Now, take $\log_{10}$
$$t = 20(1-\log_{10}8) \approx 1.94$$
A: $$20\cdot 10^{0.1t}=25\cdot 10^{0.05t}\\10^{\log(20)+0.1t}=10^{\log(25)+0.05t}\\\log(20)+0.1t=\log(25)+0.05t\\(0.1-0.05)t=\log(25)-\log(20)=\log\left(\frac{5}{4}\right)\\0.05t=\log\left(\frac{5}{4}\right)\\t=20\log\left(\frac{5}{4}\right)$$
