Question

If $lo{g}_{2}x+lo{g}_{4}x+lo{g}_{16}x=\frac{21}{4},$ then $x$ equal to

• A. 8
• B. 4
• C. 2
• D. 16

Answer 1

Answer: (A)

8

We know that ${\log }_{a^n}N=\frac{1}{n}{\log }_{a}N.$

Now,

$lo{g}_{2}x+lo{g}_{4}x+lo{g}_{16}x=\frac{21}{4}$

$\Rightarrow lo{g}_{2}x+lo{g}_{2^2}x+lo{g}_{2^4}x=\frac{21}{4}$

$\Rightarrow lo{g}_{2}x+\frac{1}{2}lo{g}_{2}x+\frac{1}{4}lo{g}_{2}x=\frac{21}{4},$

$\Rightarrow (1+\frac{1}{2}+\frac{1}{4}){\log }_{2}x=\frac{21}{4}$

$\Rightarrow \frac{7}{4}{\log }_{2}x=\frac{21}{4}$

$\Rightarrow {\log }_{2}x=3$

Hence, $x=2^3=8$