ABC26GN2061 · Logarithms

Subject: General Aptitude · Chapter: Logarithms · Exam: · Marks: · Difficulty:

The value of $\left(\log _{9} 27+\log _{8} 32\right)$ is
(a)$\frac{7}{2}$
(b)$\frac{19}{6}$
(c)4
(d)7
Answer
Answer (as printed): B
Explanation
$\log _{9} 27+\log _{8} 32=\log _{3^{2}}\left(3^{3}\right)+\log _{2^{3}}\left(2^{5}\right)$ $$=\frac{3}{2} \log _{3} 3+\frac{5}{3} \log _{2} 2=\frac{3}{2}+\frac{5}{3}=\frac{9+10}{6}=\frac{19}{6} .$$

Explanation as extracted from the printed page; notation may be imperfect.

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