Question

Solve the following system of equations.
 $ {3^x} \cdot {2^y} = 576 $ , $ {\log _{\sqrt 2 }}\left( {y - x} \right) = 4 $

Answer 1

Hint: In this question, two equations are given and we have to solve them and find the value of $ x{\rm{ and }}y $ variables. We first simplify one equation and find a relation between both the variables and then using that relationship, we can calculate the value of the variables.

Complete step-by-step answer:
The system of equations-
The first equation is,
 $ {3^x} \cdot {2^y} = 576 $
And, the second equation is,
 $ {\log _{\sqrt 2 }}\left( {y - x} \right) = 4 $
Now considering the second equation and solving it step by step, we get,
\[{\log _{\sqrt 2 }}\left( {y - x} \right) = 4\]
In order to remove the logarithm, we have to raise both sides of the equation to the same value of exponent as the base of the log value. So, removing the \[{\log _{\sqrt 2 }}\] by raising both sides to $ \sqrt 2 $ we get,
 $ $ \[\begin{array}{c}
\left( {y - x} \right) = {\left( {\sqrt 2 } \right)^4}\\
\left( {y - x} \right) = 4\\
y = 4 + x
\end{array}\]
So, now we have a relation between the variables $ x{\rm{ and }}y $ given by
 $ y = 4 + x $
Now substituting this value of $ y $ in the second equation and solving it for the values of $ x{\rm{ and }}y $ -
 $ {3^x} \cdot {2^{4 + x}} = 576 $
We can write this as-
 $ {3^x} \cdot {2^4} \cdot {2^x} = 576 $
We know that $ {a^x} \cdot {b^x} = {\left( {ab} \right)^x} $ and $ {2^4} = 16 $ , substituting this in the equation we get,
 $ \begin{array}{c}
16 \cdot {6^x} = 576\\
{6^x} = \dfrac{{576}}{{16}}\\
{6^x} = 36
\end{array} $
We can write $ 36 $ as $ 36 = {6^2} $ , putting this in the equation we get,
 $ {6^x} = {6^2} $
Comparing the exponents from both the sides we get,
 $ x = 2 $
Substituting this value of $ x $ in the first equation and solving it for $ y $ we get,
 $ \begin{array}{c}
{3^2} \cdot {2^y} = 576\\
9 \cdot {2^y} = 576\\
{2^y} = \dfrac{{576}}{9}\\
{2^y} = 64
\end{array} $
We can write $ 64 $ as $ 64 = {2^6} $ , putting this in the equation we get,
 $ {2^y} = {2^6} $
Comparing the exponents from both the sides we get,
 $ y = 6 $
Therefore, The value of $ x $ is $ 2 $ and the value of $ y $ is $ 6 $ .

Note: If we consider the first equation $ {3^x} \cdot {2^y} = 576 $ and then from this equation, we try to find a relation between variables, $ x{\rm{ and }}y $ the solution gets more complex and harder to solve. That is the reason we have considered the second equation and derived the relation between $ x{\rm{ and }}y $ from it.