Question

9^(x+1) =27/(3^x). solve for x​

Answer 1

Solution:-

$\sf \to \: {9}^{x + 1} = \dfrac{27}{ {3}^{x} }$

$\sf \to \: 3 {}^{2(x + 1)} \times {3}^{x} = 27$

$\sf \to \: {3}^{2x + 2} \times {3}^{x} = 3 {}^{3}$

$\sf \to \: {3}^{2x + 2 + x} = {3}^{3}$

$\sf \to \: {3}^{3x + 2} = {3}^{3}$

$\sf \to \: 3x + 2 = 3$

$\sf \to \: 3x = 3 - 2$

$\sf \to \: 3x = 1$

$\sf \to \: x = \dfrac{1}{3}$

Some exponential law

$\sf \to \: {a}^{m} \times a {}^{n} = {a}^{m + n}$

$\sf \to \: {a}^{0} = 1$

$\sf \to \: {a}^{ - 1} = \dfrac{1}{a}$

$\sf \to \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\sf \to \: ( {a}^{m} ) {}^{n} = {a}^{mn}$

Answer 2

3^(2x+2) = 3^(3-x)
So, 2x + 2 = 3 - x
3x = 1
x = 1/3