How do you solve 4^(3x-2)=1?

Sep 10, 2016

$x = \frac{2}{3}$

Explanation:

As any number to the power of zero is one, $\implies {4}^{0} = 1$
Since ${4}^{0} = 1 \implies {4}^{3 x - 2} = 1 , \implies 3 x - 2 = 0$
Since $3 x - 2 = 0 , \implies 3 x = 2 , \implies x = \frac{2}{3}$

Sep 10, 2016

To solve an exponential equation, take the log of each side. Or remember that anything raised to the zero power equals 1.

Explanation:

Take the log of each side.
$\log {4}^{3 x - 2} = \log 1$

By log properties, the exponent becomes a coefficient
($\log {x}^{a} = a \log x$)
$\left(3 x - 2\right) \log 4 = \log 1$

Divide both sides by log 4
$3 x - 2 = \log \frac{1}{\log} 4$

$\log 1 = 0$ so $\log \frac{1}{\log} 4 = 0$

$3 x - 2 = 0$

Subtract 2 from each side.
$3 x = 2$

Divide each side by 3
$x = \frac{2}{3}$

OR

Remember that anything raised to the zero power equals 1
(${x}^{0} = 1$)

So the exponent $3 x - 2$ must equal zero.

Solve
$3 x - 2 = 0$
$x = \frac{2}{3}$