Question

prove that
$log \frac{7}{8} + log \frac{32}{49} - log \frac{4}{14} = log2$


right answer= brainalist​

Answer 1

Answer:

See the explanation

Step-by-step explanation:

  $\displaystyle \text{LHS}=(\log \frac{7}{8}+\log \frac{32}{49})-\log\frac{4}{14}$

• Using product rule $\log a+\log b=\log(ab)$

         $\displaystyle =\log(\frac{7}{8}\times \frac{32}{49})-\log \frac{2}{7}\\\\=\log (\frac{4}{7})-\log \frac{2}{7}$  

• Using quotient rule $\log a-\log =\log \frac{a}{b}$

         $=\log( \dfrac{\frac{4}{7}}{\frac{2}{7}})=\log (\frac{4}{7}\times \frac{7}{2})=\log 2=\text{RHS}$