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

The Correct Option is A

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): \(4 < a < 5\)