Question

4) $if\frac{1}{3}{\log }_{10}\sqrt{a}-\frac{1}{2}{\log }_{10}\sqrt[3]{b}-\frac{1}{6}=0$, find a in the terms of b .

Answer 1

Dear student
$$\begin{gathered} \frac{1}{3}{\log }_{10}\sqrt{\mathrm{a}}-\frac{1}{2}{\log }_{10}\sqrt[3]{\mathrm{b}}-\frac{1}{6}=0 \\ \Rightarrow \frac{1}{3}{\log }_{10}{\left(\mathrm{a}\right)}^{\frac{1}{2}}-\frac{1}{2}{\log }_{10}{\left(\mathrm{b}\right)}^{\frac{1}{3}}-\frac{1}{6}=0 \\ \Rightarrow \frac{1}{6}{\log }_{10}\left(\mathrm{a}\right)-\frac{1}{6}{\log }_{10}\left(\mathrm{b}\right)-\frac{1}{6}=0\left[\because {\log }_{\mathrm{b}}\left({\mathrm{a}}^{\mathrm{c}}\right)={\mathrm{clog}}_{\mathrm{b}}\left(\mathrm{a}\right)\right] \\ \Rightarrow {\log }_{10}\left(\mathrm{a}\right)-{\log }_{10}\left(\mathrm{b}\right)-1=0 \\ \Rightarrow {\log }_{10}\left(\mathrm{a}\right)={\log }_{10}\left(\mathrm{b}\right)+1 \\ \Rightarrow {\log }_{10}\left(\mathrm{a}\right)={\log }_{10}\left(\mathrm{b}\right)+{\log }_{10}\left(10\right) \\ \Rightarrow {\log }_{10}\left(\mathrm{a}\right)={\log }_{10}\left(10\mathrm{b}\right) \\ \Rightarrow \mathrm{a}=10\mathrm{b} \end{gathered}$$
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