Question

If the values $ {{4}^{x}} $ , $ {{8}^{y}} $ and $ {{16}^{z}} $ follows the condition $ {{4}^{x}}={{8}^{y}}={{16}^{z}} $ , then prove that $ \dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z} $ .

Answer 1

Hint: We take the equalities and convert 4, 8 and 16 in powers of 2. We use conditions like $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ and if $ {{a}^{m}}={{a}^{n}} $ , then $ m=n $ to get ‘x’ and ‘z’ in terms of ‘y’. We take the R.H.S (Right Hand Side) of $ \dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z} $ and substitute the values of ‘x’ and ‘z’ obtained in terms of ‘y’ to get the required proof.

Complete step-by-step answer:
Given that we had a condition $ {{4}^{x}}={{8}^{y}}={{16}^{z}} $ that holds true. We need to prove $ \dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z} $ .
Let us consider $ {{4}^{x}}={{8}^{y}} $ ---(1).
We know that $ 4={{2}^{2}} $ and $ 8={{2}^{3}} $ . Let us substitute these in place of ‘4’ and ‘8’ of equation (1).
So, we have $ {{\left( {{2}^{2}} \right)}^{x}}={{\left( {{2}^{3}} \right)}^{y}} $ ---(2).
We know that $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ . We use this result in equation (2).
So, $ {{2}^{2\times x}}={{2}^{3\times y}} $ .
 $ {{2}^{2x}}={{2}^{3y}} $ ---(3).
We know that if $ {{a}^{m}}={{a}^{n}} $ , then $ m=n $ . We use this result in equation (3).
We get 2x=3y.
 $ x=\dfrac{3y}{2} $ ---(4).
Let us consider $ {{8}^{y}}={{16}^{z}} $ ---(5).
We know that $ 8={{2}^{3}} $ and $ 16={{2}^{4}} $ . Let us substitute these in place of ‘8’ and ‘16’ of equation (5).
So, we have $ {{\left( {{2}^{3}} \right)}^{y}}={{\left( {{2}^{4}} \right)}^{z}} $ ---(6).
We know that $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ . We use this result in equation (6).
So,\[{{2}^{3\times y}}={{2}^{4\times z}}\].
 $ {{2}^{3y}}={{2}^{4z}} $ ---(7).
We know that if $ {{a}^{m}}={{a}^{n}} $ , then $ m=n $ . We use this result in equation (7).
We get 3y=4z.
 $ z=\dfrac{3y}{4} $ ---(8).
Now we take the R.H.S (Right Hand Side) of $ \dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z} $ to proceed through the problem.
Let us consider $ \dfrac{1}{x}+\dfrac{1}{z} $ ---(9).
We substitute the values of ‘x’ and ‘z’ which were obtained from equations (4) and (8) in equation (9).
 $ \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{\dfrac{3y}{2}}+\dfrac{1}{\dfrac{3y}{4}} $ .
 $ \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{3y}+\dfrac{4}{3y} $ .
 $ \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2+4}{3y} $ .
 $ \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{6}{3y} $ .
 $ \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{y} $ ---(10).
From equation (10), we can see that we have proved $ \dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z} $ .

Note: Alternatively, we can solve the problem as shown below:
Let us assume $ {{4}^{x}}={{8}^{y}}={{16}^{z}}=r $ (say).
We have $ {{4}^{x}}=r $ . We know that if $ {{a}^{x}}=y $ , then $ x={{\log }_{a}}\left( y \right) $ .
So, we get $ x={{\log }_{4}}\left( r \right) $ .
 $ x={{\log }_{{{2}^{2}}}}\left( r \right) $ .
We know that $ {{\log }_{{{a}^{n}}}}\left( y \right)=\dfrac{1}{n}{{\log }_{a}}\left( y \right) $ .
So, we get $ x=\dfrac{1}{2}{{\log }_{2}}\left( r \right) $ ---(11).
We have $ {{8}^{y}}=r $ . We know that if $ {{a}^{x}}=y $ , then $ x={{\log }_{a}}\left( y \right) $ .
So, we get $ y={{\log }_{8}}\left( r \right) $ .
 $ y={{\log }_{{{2}^{3}}}}\left( r \right) $ .
We know that $ {{\log }_{{{a}^{n}}}}\left( y \right)=\dfrac{1}{n}{{\log }_{a}}\left( y \right) $ .
So, we get $ y=\dfrac{1}{3}{{\log }_{2}}\left( r \right) $ ---(12).
We have $ {{16}^{z}}=r $ . We know that if $ {{a}^{x}}=y $ , then $ x={{\log }_{a}}\left( y \right) $ .
So, we get $ z={{\log }_{16}}\left( r \right) $ .
 $ z={{\log }_{{{2}^{4}}}}\left( r \right) $ .
We know that $ {{\log }_{{{a}^{n}}}}\left( y \right)=\dfrac{1}{n}{{\log }_{a}}\left( y \right) $ .
So, we get $ z=\dfrac{1}{4}{{\log }_{2}}\left( r \right) $ ---(13).
Let us consider $ \dfrac{1}{x}+\dfrac{1}{z} $ and substitute the values of ‘x’ and ‘z’ obtained from equations (11) and (13).
 $ \dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{\dfrac{{{\log }_{2}}r}{2}}+\dfrac{1}{\dfrac{{{\log }_{2}}r}{4}} $ .
\[\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{{{\log }_{2}}r}+\dfrac{4}{{{\log }_{2}}r}\].
\[\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{6}{{{\log }_{2}}r}\].
\[\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2\times 3}{{{\log }_{2}}r}\].
\[\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{\dfrac{{{\log }_{2}}r}{3}}\].
From equation (12), we get
\[\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{2}{y}\].
We have proved $ \dfrac{2}{y}=\dfrac{1}{x}+\dfrac{1}{z} $ .