4. A question on logarithms (Eton College 01C_MT1 Q16)

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This question comes from a test paper for A-level maths for Michaelmas Term at Eton College.

Q16. (a) Solve the equation $$ \begin{align} \log_{10}\left(3^x-5^{2-x}\right)=2+\log_{10}(2)-x\log_{10}(5), \end{align} $$ giving your answer as exact. [3 marks]

(b) Solve the equation $$ \begin{align} \left(5\sqrt{5}\right)^{3x}-31=30\left(\frac1{\sqrt[4]{5}}\right)^{9x}, \end{align} $$ giving your answer correct to 3 significant figures. [4 marks]

 

Solution. (a) We find $$ \begin{align} \log_{10}\left(3^x-5^{2-x}\right)&=2+\log_{10}(2)-x\log_{10}(5) \\ &=\log_{10}\left(10^2\right)+\log_{10}(2)-\log_{10}\left(5^x\right) \\ &=\log_{10}\left(\frac{200}{5^x}\right) \end{align} $$ Since logarithm is a one-to-one function, this gives $$ \begin{align} \Rightarrow\quad 3^x-5^{2-x}&=\frac{200}{5^x} \\ \Rightarrow\quad 3^x-\frac{25}{5^x}&=\frac{200}{5^x} \\ \Rightarrow\quad 3^x\times 5^x-25&=200 \\ \Rightarrow\quad 15^x&=225 \\ \\ \Rightarrow\quad x&=\log_{15}(225) \\ &=\log_{15}\left(15^2\right) \\ &=2 \end{align} $$

 

(b) We have $$ \begin{align} \left(5\sqrt{5}\right)^{3x}-31&=30\left(\frac1{\sqrt[4]{5}}\right)^{9x} \\ \Rightarrow\quad \left(5^{\frac32}\right)^{3x}-31&=30\left(5^{-\frac14}\right)^{9x} \\ \Rightarrow\quad 5^{\frac{9x}{2}}-31&=30\left({5^{-\frac{9x}{4}}}\right) \end{align} $$ We let $t=5^{\frac{9x}{4}}$, then the equation reads $$ \begin{align} t^2-31&=\frac{30}{t} \\ \Rightarrow\quad f(t)=t^3-31t-30&=0 \end{align} $$ We note that $f(-1)=0$ and, by the factor theorem, $(t+1)$ is a factor, i.e. $$ \begin{align} f(t)&=(t+1)\left(t^2-t-30\right) \\ &=(t+1)(t+5)(t-6) \\ \\ \Rightarrow\quad t&=-1,\quad -5,\quad 6 \end{align} $$ Since $t=5^{\frac{9x}{4}}>0$ for real values $x$, only $t=6$ gives a real solution for $x$. $$ \begin{align} t=5^{\frac{9x}{4}}&=6 \\ \Rightarrow\quad \frac{9x}{4}&=\log_5(6) \\ \Rightarrow\quad x&=\frac49\log_5(6)=0.495\quad{\rm (3\;s.f.)} \end{align} $$