Question

If ${4^x} = {7^y} = {112^z}$ then prove that $\dfrac{2}{x} + \dfrac{1}{y} = \dfrac{1}{z}$

Answer 1

Hint: We will use logarithm to simplify this question. then, find the value of $x$ and $y$ in terms of $z$ from the equation. Then, substitute these values in the left-hand-side of the equation which have to prove. Apply the properties of logarithm, $\log {m^n} = n\log m$ and $\log a + \log b = \log \left( {ab} \right)$ to make the expression equal to RHS.

Complete step-by-step answer:
We are given that ${4^x} = {7^y} = {112^z}$.
Whenever we have an expression, where the variable is in power, we simplify the expression by taking log on both sides because $\log {m^n} = n\log m$
First let, ${4^x} = {7^y} = {112^z}$
Take a log on all sides.
$x\log 4 = y\log 7 = z\log 112$ eqn. (1)
We have to prove $\dfrac{2}{x} + \dfrac{1}{y} = \dfrac{1}{z}$.
We will find the value of $x$ and $y$ in terms of $z$ from equation (1).
Now,
 $
  x\log 4 = z\log 112 \\
   \Rightarrow x = \dfrac{{z\log 112}}{{\log 4}} \\
$
Similarly,
$
  y\log 7 = z\log 112 \\
   \Rightarrow y = z\dfrac{{\log 112}}{{\log 7}} \\
 $
Now, on substituting the values in the LHS of the equation $\dfrac{2}{x} + \dfrac{1}{y} = \dfrac{1}{z}$, we get,
$\dfrac{2}{{\dfrac{{z\log 112}}{{\log 4}}}} + \dfrac{1}{{\dfrac{{z\log 112}}{{\log 7}}}} = \dfrac{1}{z}$
The above equation can also be written as
$
  \dfrac{{2\log 4}}{{z\log 112}} + \dfrac{{\log 7}}{{z\log 112}} \\
   \Rightarrow \dfrac{{2\log 4 + \log 7}}{{z\log 112}} \\
$
As we know that $\log {m^n} = n\log m$, we will take 2 that is multiplied with log4 in the power of 4.
$
  \dfrac{{\log {{\left( 4 \right)}^2} + \log 7}}{{z\log 112}} \\
   \Rightarrow \dfrac{{\log 16 + \log 7}}{{z\log 112}} \\
$
Also, $\log a + \log b = \log \left( {ab} \right)$.
Thus, we will simplify the expression in the numerator using the above property.
$
  \dfrac{{\log \left( {16 \times 7} \right)}}{{z\log 112}} \\
   \Rightarrow \dfrac{{\log 112}}{{z\log 112}} \\
   \Rightarrow \dfrac{1}{z} \\
$
Thus, LHS=RHS.
Hence, proved.

Note: In these types of we use properties of log to simplify the expression. The inverse function of exponentiation is logarithm. One should know all the properties of logarithm to solve this question correctly.