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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
- $6$
- $7$

## 1 Answer

1 vote

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 $:\text{B}$

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 $:\text{B}$