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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

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Taking logs of both sides, then splitting the products into sums by logarithm rules, yields $$\log(20\cdot10^{0.1t})=\log(25\cdot10^{0.05t})$$ $$\log20+0.1t=\log25+0.05t$$ $$0.05t=\log25-\log20=\log\frac54$$ $$t=20\log\frac54=\log\frac{5^{20}}{4^{20}}$$

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  • $\begingroup$ hey sir thank you, may i ask you how did you get the 20 in $20log5/4$ $\endgroup$ – Fred Weasley Nov 5 '18 at 11:50
  • $\begingroup$ @Tfue By multiplying by $\frac1{0.05}=20$ in the third line. $\endgroup$ – Parcly Taxel Nov 5 '18 at 11:51
  • $\begingroup$ AH i see sir, thank you very much! $\endgroup$ – Fred Weasley Nov 5 '18 at 11:52
  • $\begingroup$ sorry sir 1 more question, in the second line how did you get the +0.1t and the +0.05t ? I thought you have to times? wouldn't be $log_{10} 10^{0.1t}$ equal to x0.1t instead of +0.1t ? $\endgroup$ – Fred Weasley Nov 5 '18 at 12:00
  • $\begingroup$ @Tfue $\log(20\cdot10^{0.1t})=\log20+\log10^{0.1t}=\log20+0.1t$, because $\log10^x=x$. Same for the RHS. $\endgroup$ – Parcly Taxel Nov 5 '18 at 12:01
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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.

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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 $\log_{10}$ $$t = 20(1-\log_{10}8) \approx 1.94$$

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  • $\begingroup$ awesome thanks! very insightful! $\endgroup$ – Fred Weasley Nov 5 '18 at 11:53
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$$20\cdot 10^{0.1t}=25\cdot 10^{0.05t}\\10^{\log(20)+0.1t}=10^{\log(25)+0.05t}\\\log(20)+0.1t=\log(25)+0.05t\\(0.1-0.05)t=\log(25)-\log(20)=\log\left(\frac{5}{4}\right)\\0.05t=\log\left(\frac{5}{4}\right)\\t=20\log\left(\frac{5}{4}\right)$$

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