Question

If ${\log }_{2}x+{\log }_{4}x+{\log }_{64}x=5$, then the value of $x$ will be

• A. $8$
• B. $16$
• C. $7$
• D. $2$

Answer 1

The correct option is A $8$

${\log }_{2}x+{\log }_{4}x+{\log }_{64}x=5$

We know that,

${\log }_{b}a=\frac{{\log }_{c}a}{{\log }_{c}b}$

$\therefore \frac{\log x}{\log 2}+\frac{\log x}{\log 4}+\frac{\log x}{\log 64}=5$

Assuming base $2$ for numerator and denominator, we get

$\frac{\log x}{1}+\frac{\log x}{2}+\frac{\log x}{6}=5$

$6\log x+3\log x+\log x=30$

$10\log x=30$

${\log }_2{x}^{10}=30\dots ({\log }_y{x}^n=n{\log }_{y}x)$

$x^{10}=2^{30}$

So, $x=2^3=8$