1 1 vote If $\log\left ( 2^{a} \times 3^{b}\times 5^{c}\right )$ is the arithmetic mean of $\log\left ( 2^{2} \times 3^{3}\times 5 \right ),$ $\log\left ( 2^{6} \times3\times 5^{7} \right ),$ and $\log\left ( 2 \times3^{2}\times 5^{4} \right ),$ then $a$ equals $2$ None of these $6$ $7$ Quantitative Aptitude cat2017-2 quantitative-aptitude arithmetic-progression arithmetic-mean + – go_editor 14.2k points 2.0k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
5 5 votes We know that the arithmetic mean of $x, y,$ and $z$ is $\frac{x+y+z}{3}.$ Now, $\log(2^{a} \times 3^{b} \times 5^{c}) = \frac{\log\left(2^{2} \times 3^{3} \times 5\right)+\log\left(2^{6} \times 3 \times 5^{7}\right)+\log\left(2 \times 3^{2} \times 5^{4}\right)}{3}$ $\Rightarrow \log(2^{a} \times 3^{b} \times 5^{c}) = \frac{\log\left(2^{2} \times 3^{3} \times 5 \times 2^{6} \times 3 \times 5^{7} \times 2 \times 3^{2} \times 5^{4}\right)}{3}$ $\Rightarrow \log(2^{a} \times 3^{b} \times 5^{c}) = \frac{1}{3}\log(2^{9} \times 3^{6} \times 5^{12}) \quad [\because \log_{b}{m}+\log_{b}{n}=\log_{b}{mn}]$ $\Rightarrow \log(2^{a} \times 3^{b} \times 5^{c}) = \frac{1}{3}\log(2^{3} \times 3^{2} \times 5^{4})^{3} $ $\Rightarrow \log(2^{a} \times 3^{b} \times 5^{c}) =\frac{3}{3}\log(2^{3} \times 3^{2} \times 5^{4}) \quad [\because \log_{b}a^{x} = x \log_{b}a] $ $\Rightarrow \log(2^{a} \times 3^{b} \times 5^{c})= \log(2^{3} \times 3^{2} \times 5^{4}) $ If two terms have the same base, then we can equate their powers, we get $\boxed{a = 3}.$ Correct Answer $:\text{B}$ Anjana5051 answered Dec 31, 2021 • edited Dec 31, 2021 by Lakshman Bhaiya Anjana5051 12.0k points comment Share Follow 0 reply Please log in or register to add a comment.