Question

solve it..value of x

Answer 1

Given :-

${3}^{2x + 4} + 1 = 2 \times {3}^{x + 2}$


we can write 3^2x as
${3}^{2x} = ({3}^{x} ) {}^{2}$


and,
${3}^{2x + 4} = {3}^{2x} \times {3}^{4} \\ \\ = > ({3}^{x} ) {}^{2} \times {3}^{4}$


So,

${3}^{2x + 4} + 1 = 2 \times {3}^{x + 2} \\ \\ = > ( {3}^{x} ) {}^{2} \times {3}^{4} + 1 = 2 \times {3}^{x} \times 3 {}^{2} \\ \\ = > ( {3}^{x} ) {}^{2} \times 81 + 1 = 2 \times {3}^{x} \times 9 \\ \\ = > ( {3}^{x} ) {}^{2} \times 81 + 1 = 18 \times {3}^{x}$
$let \: {3}^{x} = y$
$so \\ ( {3}^{x} ) {}^{2} \times 81 + 1 = 18 \times {3}^{x} \\ \\ = > 81{y}^{2} + 1 = 18y \\ \\ = > 81 {y}^{2} - 18y + 1 = 0$


Now we have,

81y² - 18y + 1 = 0

By splitting the middle term,

81y² - 9y - 9y + 1 = 0

=> 9y(9y - 1) - 1(9y - 1) = 0

=> (9y - 1)(9y - 1) = 0

=> (9y - 1)² = 0

=> 9y - 1 = 0

=> 9y = 1

=> y = 1/9

Now we have assumed that,
$y = {3}^{x}$

But y = 1/9

$= > {3}^{x} = \frac{1}{9} \\ \\ = > {3}^{x} = {3}^{ - 2} \\ \\ = > x = - 2$


So the value of x is -2


Hope it helps dear friend ☺️