Question

If $ {{11}^{x}}={{3}^{y}}={{99}^{z}}, $ then $ \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=--. $
(a) $ \dfrac{2}{z}-\dfrac{1}{y} $
(b) $ \dfrac{2}{z}+\dfrac{1}{y} $
(c) $ -\dfrac{1}{y} $
(d)0

Answer 1

Hint: To solve the question given above, we will equate the relation $ {{11}^{x}}={{3}^{y}}={{99}^{z}} $ to ‘k’ and then we will find the values of x, y and z in terms of k. After finding out these values, we will put them in $ \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} $ . We will assume that its result will be ‘p’. Now, we will check each option one by one and the option which will give value equal to ‘p’ will be the answer of this question.

Complete step-by-step answer:
To start with, we will equate the relation $ {{11}^{x}}={{3}^{y}}={{99}^{z}} $ to ‘k’. Thus, we have the following relation:
 $ {{11}^{x}}={{3}^{y}}={{99}^{z}}=k.......\left( 1 \right) $
From (1), we can say that:
 $ {{11}^{x}}=k $
We will take logarithms on both sides. Thus, we will get:
 $ \log {{11}^{x}}=\log k $
In the above equation, we will use a logarithmic identity:
 $ \log {{x}^{y}}=y\log x $
Thus, after applying the above identity, we will get following equation:
 $ x\log 11=\log k $
 $ \Rightarrow \dfrac{1}{x}=\dfrac{\log 11}{\log k}......\left( 2 \right) $
From (1), we can also say that:
 $ {{3}^{y}}=k $
On taking logarithm on both sides we will get:
  $ \log {{3}^{y}}=\log k $
 $ y\log 3=\log k $
 $ \Rightarrow \dfrac{1}{y}=\dfrac{\log 3}{\log k}.........\left( 3 \right) $
Similarly, we can say that: $ \dfrac{1}{z}=\dfrac{\log 99}{\log k}.........\left( 4 \right) $
Now, we will assume that the value of $ \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} $ is ‘p’.
Thus, $ p=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}......\left( 5 \right) $
From (2), (3) and (4), we will put the values of $ \dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z} $ into equation (5). After doing this we will get:
 $ p=\dfrac{\log 11}{\log k}+\dfrac{\log 3}{\log k}+\dfrac{\log 99}{\log k} $
 $ p=\dfrac{\log 11+\log 3+\log 99}{\log k} $
Now, we will use another logarithmic identity:
 $ \log a+\log b+\log c=\log \left( abc \right) $
 $ \Rightarrow p=\dfrac{\log \left( 11\times 3\times 99 \right)}{\log k} $
 $ \Rightarrow p=\dfrac{\log \left( 3267 \right)}{\log k} $
Now, we will check the options one by one:
Option (a): $ \dfrac{2}{z}-\dfrac{1}{y}=\dfrac{2\log 99}{\log k}-\dfrac{\log 3}{\log k} $
 $ =\dfrac{\log {{99}^{2}}}{\log k}-\dfrac{\log 3}{\log k} $
Now, we will use the following identity:
 $ \log a-\log b=\log \left( \dfrac{a}{b} \right) $
 $ \dfrac{2}{z}-\dfrac{1}{y}=\dfrac{\log \left( \dfrac{99\times 99}{3} \right)}{\log k}=\dfrac{\log \left( 3267 \right)}{\log k}=p $
Option (b): $ \dfrac{2}{z}+\dfrac{1}{y}=\dfrac{2\log 99}{\log k}+\dfrac{\log 3}{\log k}=\dfrac{\log {{\left( 99 \right)}^{2}}}{\log k}+\dfrac{\log 3}{\log k} $
 $ =\dfrac{\log \left( 99\times 99\times 3 \right)}{\log k}=\dfrac{\log \left( 29403 \right)}{\log k}\ne p $
Option (c): $ \dfrac{-1}{y}=\dfrac{-1\times \log 3}{\log k}\ne p $
Option (d): $ 0\ne p $
Hence, option (a) is correct.

Note: The above question can also be solved alternatively by the following method:
 $ {{11}^{x}}={{3}^{y}}={{99}^{z}}=k $
 $ \Rightarrow 11=\sqrt[x]{k}........\left( 1 \right) $
 $ \Rightarrow 3=\sqrt[y]{k}........\left( 2 \right) $
 $ \Rightarrow 99=\sqrt[z]{k}........\left( 3 \right) $
 $ 99=3\times 3\times 11 $
 $ \Rightarrow \sqrt[z]{k}=\sqrt[y]{k}\times \sqrt[y]{k}\times \sqrt[x]{k} $
 $ \Rightarrow {{\left( k \right)}^{\dfrac{1}{z}}}={{\left( k \right)}^{\dfrac{2}{y}}}\times {{\left( k \right)}^{\dfrac{1}{x}}} $
 $ \Rightarrow \dfrac{1}{z}=\dfrac{2}{y}+\dfrac{1}{x}\Rightarrow \dfrac{1}{x}=\dfrac{1}{z}-\dfrac{2}{y} $
Thus, the value of $ \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{z}-\dfrac{2}{y}+\dfrac{1}{y}+\dfrac{1}{z} $
 $ \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{2}{z}-\dfrac{1}{y} $ .