# How do you solve  8^(6x-5)=(1/16)^(6x-2)?

Mar 3, 2016

x =$\frac{23}{42}$

#### Explanation:

${a}^{m + n} = {a}^{m} . {a}^{n}$
Express 8 and 16 in powers of 2. cross multiply.
${2}^{42 x - 23}$ = 1.
${a}^{0} = 1$.
$42 x - 23$=0..

Mar 3, 2016

$x = \frac{23}{42}$

#### Explanation:

We will write $8$ and $\frac{1}{16}$ as powers of $2 :$

$8 = {2}^{3}$

$\frac{1}{16} = \frac{1}{2} ^ 4 = {2}^{-} 4$

The equation can then be rewritten as

${\left({2}^{3}\right)}^{6 x - 5} = {\left({2}^{-} 4\right)}^{6 x - 2}$

We then use the rule:

${\left({a}^{b}\right)}^{c} = {a}^{b c}$

This makes the equation

${2}^{3 \left(6 x - 5\right)} = {2}^{- 4 \left(6 x - 2\right)}$

Since the bases are equal, the exponents can be set equal to one another as well:

$3 \left(6 x - 5\right) = - 4 \left(6 x - 2\right)$

$18 x - 15 = - 24 x + 8$

$42 x = 23$

$x = \frac{23}{42}$