Question

For a real number a, if $\frac{log_{15}a+log_{32}a}{(log_{15}a)(log_{32}a)}= 4$, then a must lie in the range

• A. $4 < a < 5$
• B. $3 < a < 4$
• C. $a > 5$
• D. $2 < a < 3$

Answer 1

Answer: (A)

$4 < a < 5$

Answer 2

Given: $\frac{log_{15}a+log_{32}a}{(log _{15}a)(log_{32}a)}=4$

$⇒ \frac{\left(\frac{loga}{log15}+\frac{loga}{log32}\right)}{\frac{loga}{log15} \times \frac{ loga}{log32} }=4$

On solving the above equation,
$log\space a(log32 +log 15)=4(log a)^ 2$
$log\space a(log32 +log 15)=log \ a\ .\ 4\ log a$
$log32 +log 15)=4\ log a$
$log\ (32\times 15)=loga^ 4$
$log\ 480=loga^ 4$
$⇒a^ 4 =480$
We know that,
$4^4=256$
$5^4=625$
$⇒ a$ is between $4$ and $5$.

So, the correct option is (A): $4 < a < 5$