How do you solve 6^ { 2x + 3} \cdot 6^ { 1- x } = \frac { 1} { 6}?

May 4, 2017

$x = - 3$

Explanation:

convert $\frac{1}{6}$ to a power of $6$:

$\frac{1}{a} ^ m = {a}^{-} m$

$\frac{1}{6} ^ 1 = {6}^{-} 1$

this gives you:

${6}^{2 x + 3} \cdot {6}^{1 - x} = {6}^{-} 1$

law of indices:
${a}^{m} \cdot {a}^{n} = {a}^{m + n}$

using this:

${6}^{2 x + 3} \cdot {6}^{1 - x} = {6}^{2 x + 3 + 1 - x} = {6}^{-} 1$

this gives the equation
$2 x + 3 + 1 - x = - 1$

collect like terms:
$x + 4 = - 1$

subtract $4$:

$x = - 3$