Question

If ${\log }_{4}32=x-4$, then the least positive integer greater than $x$ is

• A. $6$
• B. $7$
• C. $8$
• D. $5$

Answer 1

The correct option is B $7$

${\log }_{4}32=x-4$

$\therefore \frac{\log 32}{\log 4}=x-4$ ....(${\log }_{b}a=\frac{\log a}{\log b}$)

$\therefore \frac{\log 2^5}{\log 2^2}=x-4$

$\therefore x-4=\frac{5}{2}$ $\Rightarrow x=6\frac{1}{2}$

Integer just greater than $x=6\frac{1}{2}$ is $7$.

(Whenever you see numbers inside the logarithm in some powers, try to simplify them by taking powers out, that may boil down into something easily solvable.)