Logarithm

[old prawn]

someone

[hasty barn]

Can you express 0.001 and 0.1 as powers of 10

Think about the negative powers of 10

@fluid yacht

[rotund canopy]

\begin{align*} \frac{1}{2}\big(\log_{10}{x}\big)\big(\log_{10}{0.001x}\big) &= \log_{10}{0.1} \ \frac{1}{2}\big(\log_{10}{x}\big)\big(\log_{10}{0.001} + \log_{10}{x}\big) &= \log_{10}{0.1} \qquad \text{using log laws} \ \frac{1}{2}\big(\log_{10}{x}\big)\big(\log_{10}{10^{-3}} + \log_{10}{x}\big) &= \log_{10}{10^{-1}} \qquad \text{using index laws} \ \text{noting that $0.001 = \frac{1}{1000} = 10^{-3}$ and}&\ \text{that $0.1=\frac{1}{10} = 10^{-1}$} \ \frac{1}{2}\big(\log_{10}{x}\big)\big(-3\log_{10}{10} + \log_{10}{x}\big) &= -\log_{10}{10} \qquad \text{using log laws} \ \frac{1}{2}\big(\log_{10}{x}\big)\big(-3 + \log_{10}{x}\big) &= -1 \qquad \text{noting that $\log_{10}{10}=1$} \end{align*}
This is a quadratic equation in $\log_{10}{x}$, which can be solved,

recognising that the domain of possible solutions is $x > 0$.