assume u=v=w=10

we will get ans as 1

we will get ans as 1

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The value of the expression $\dfrac{1}{1+ \log_u \: vw} + \dfrac{1}{1+ \log_v \: wu} + \dfrac{1}{1+\log_w uv}$ is _______

- $-1$
- $0$
- $1$
- $3$

Migrated from GO Mechanical 1 year ago by Arjun

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Best answer

$\dfrac{1}{1+ \log_{u} \: vw} + \dfrac{1}{1+ \log_{v} \: wu} + \dfrac{1}{1+\log_{w} uv}$

$ = \dfrac{1}{\log_{u} \: u + \log_{u} \: vw} + \dfrac{1}{\log_{v} \: v + \log_{v} \: wu} + \dfrac{1}{\log_{w} \: w +\log_{w} \: uv}$

$ = \dfrac{1}{\log_{u} \: uvw} + \dfrac{1}{\log_{v} \: vwu} + \dfrac{1}{\log_{w} \: wuv}$

$ = \dfrac{1}{\log_{u} \: uvw} + \dfrac{1}{\log_{v} \: uvw} + \dfrac{1}{\log_{w}\: uvw}$

$ = \log_{uvw} \: u + \log_{uvw} \: v + \log_{uvw} \: w$

$ = \log_{uvw} \: uvw $

$ = 1\qquad\because\left(\log_{a} \: a = 1 \right)$

Hence $(C)$ is Correct.

$ = \dfrac{1}{\log_{u} \: u + \log_{u} \: vw} + \dfrac{1}{\log_{v} \: v + \log_{v} \: wu} + \dfrac{1}{\log_{w} \: w +\log_{w} \: uv}$

$ = \dfrac{1}{\log_{u} \: uvw} + \dfrac{1}{\log_{v} \: vwu} + \dfrac{1}{\log_{w} \: wuv}$

$ = \dfrac{1}{\log_{u} \: uvw} + \dfrac{1}{\log_{v} \: uvw} + \dfrac{1}{\log_{w}\: uvw}$

$ = \log_{uvw} \: u + \log_{uvw} \: v + \log_{uvw} \: w$

$ = \log_{uvw} \: uvw $

$ = 1\qquad\because\left(\log_{a} \: a = 1 \right)$

Hence $(C)$ is Correct.

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