Question

$4 \cdot {2^{2x}} - {6^x} = 18 \cdot {3^{2x}}$

Answer 1

Hint: - Use ${6^x} = {2^x} \cdot {3^x}$
Given equation is
$4 \cdot {2^{2x}} - {6^x} = 18 \cdot {3^{2x}}$
Substitute, ${6^x} = {\left( {2 \cdot 3} \right)^x} = {2^x} \cdot {3^x}$in the above equation
$4 \cdot {2^{2x}} - {6^x} = 18 \cdot {3^{2x}} \\\ 4 \cdot {\left( {{2^x}} \right)^2} - {2^x} \cdot {3^x} = 18.{\left( {{3^x}} \right)^2} \\\ 4 \cdot {\left( {{2^x}} \right)^2} - {2^x} \cdot {3^x} - 18.{\left( {{3^x}} \right)^2} = 0 \\\$
Now factorize the above equation
$4 \cdot {\left( {{2^x}} \right)^2} - {2^x} \cdot {3^x} - 18.{\left( {{3^x}} \right)^2} = 0 \\\ 4 \cdot {\left( {{2^x}} \right)^2} + 8 \cdot {2^x} \cdot {3^x} - 9 \cdot {2^x} \cdot {3^x} - 18.{\left( {{3^x}} \right)^2} = 0 \\\ 4 \cdot {2^x}\left( {{2^x} + 2 \cdot {3^x}} \right) - 9 \cdot {3^x}\left( {{2^x} + 2 \cdot {3^x}} \right) = 0 \\\ \left( {{2^x} + 2 \cdot {3^x}} \right)\left( {4 \cdot {2^x} - 9 \cdot {3^x}} \right) = 0 \\\ \therefore \left( {{2^x} + 2 \cdot {3^x}} \right) = 0,{\text{ }}\left( {4 \cdot {2^x} - 9 \cdot {3^x}} \right) = 0 \\\$
From here, $\left( {{2^x} + 2 \cdot {3^x}} \right) = 0$ cannot be possible for any finite value of $x$.
Therefore $\left( {4 \cdot {2^x} - 9 \cdot {3^x}} \right) = 0$
$\therefore 4 \cdot {2^x} = 9 \cdot {3^x} \\\ {\left( {\dfrac{2}{3}} \right)^x} = \dfrac{9}{4} = {\left( {\dfrac{3}{2}} \right)^2} = {\left( {\dfrac{2}{3}} \right)^{ - 2}} \\\$
So on comparing the value of $x = - 2$.
So, the required solution of the given equation is $x = - 2$

Note: - Whenever we face such types of questions first convert the equation into simplified form then factorize the equation then put all the factors equal to zero and calculate the value of $x$, which is the required solution of the given equation.