Question

Prove that $7\log \left(\frac{16}{15}\right)+5\log \left(\frac{25}{24}\right)+3\log \left(\frac{81}{80}\right)=\log 2$

Answer 1

Determine the proving of the given expression $7\log \left(\frac{16}{15}\right)+5\log \left(\frac{25}{24}\right)+3\log \left(\frac{81}{80}\right)=\log 2$

Solve the L.H.S part:

$$\begin{gathered} 7\log \left(\frac{16}{15}\right)+5\log \left(\frac{25}{24}\right)+3\log \left(\frac{81}{80}\right)=7\log (\frac{2^4}{3\times 5})+5\log (\frac{5^2}{2^3\times 3})+3\log (\frac{3^4}{2^4\times 5}) \\ \Rightarrow =\log (\frac{2^{28}}{3^7\times 5^7})+\log (\frac{5^{10}}{2^{15}\times 3^5})+\log (\frac{3^{12}}{2^{12}\times 5^3})(Sincen\log m=\log m^n) \\ \Rightarrow =\log (\frac{2^{28}\times 5^{10}\times 3^{12}}{3^7\times 5^7\times 2^{15}\times 3^5\times 2^{12}\times 5^3})(Using\log a+\log b+\log c=\log abc) \\ \Rightarrow =\log \left(\frac{2^{28}\times 5^{10}\times 3^{12}}{5^{10}\times 2^{27}\times 3^{12}}\right)(Since{a}^m\times a^n=a^{m+n}) \\ \Rightarrow =\log \left(2^{28-27}\right)(Since\frac{a^m}{a^n}=a^{m-n}) \\ \Rightarrow =\log 2 \end{gathered}$$

Hence, the L.H.S=R.H.S.