Question

If $a={\log }_{2}3,b={\log }_{2}5andc={\log }_{7}2$, then ${\log }_{140}63$ in terms of $a,b$and $c$ is equal to

• A. $\frac{[2ac+1]}{[2c+abc+1]}$
• B. $\frac{[2ac+1]}{[2a+c+a]}$
• C. $\frac{[2ac+1]}{[2c+ab+a]}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Step 1. Find the value of $\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{140}}\mathbf{63}$ in terms of $\mathit{a}\mathbf{,}\mathbf{}\mathit{b}\mathbf{}$and $\mathit{c}$:

Given: $a={\log }_{2}3,b={\log }_{2}5andc={\log }_{7}2$,

${\log }_{140}63$ can be written as ${\log }_{2\times 2\times 5\times 7}3\times 3\times 7$.

Step 2. Convert $\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{2}\mathbf{\times }\mathbf{2}\mathbf{\times }\mathbf{5}\mathbf{\times }\mathbf{7}}\mathbf{3}\mathbf{\times }\mathbf{3}\mathbf{\times }\mathbf{7}$ into $\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{2}}$ by changing the base formula.

$$\begin{gathered} =\frac{{\log }_{2}3\times 3\times 7}{{\log }_{2}2\times 2\times 5\times 7}\left[\because {\log }_a\left(b\right)=\frac{{\log }_a\left(b\right)}{{\log }_2\left(a\right)}\right] \\ =\frac{{\log }_{2}3+{\log }_{2}3+{\log }_{2}7}{2{\log }_{2}2+{\log }_{2}5+{\log }_{2}7}\left[\mathbf{\because }\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathit{b}\right] \\ =\frac{2a+{\displaystyle \frac{{\log }_{7}7}{{\log }_{7}2}}}{2+b+\frac{{\log }_{7}7}{{\log }_{7}2}} \\ =\frac{2a+{\displaystyle \frac{1}{c}}}{2+b+\frac{1}{c}} \\ =\frac{\mathbf{2}\mathbf{ac}\mathbf{+}\mathbf{1}}{\mathbf{2}\mathbf{c}\mathbf{+}\mathbf{bc}\mathbf{+}\mathbf{1}} \end{gathered}$$

Hence, the option$\mathbf{\left(}\mathit{D}\mathbf{\right)}$ is correct.