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

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

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Answer 1 · 2016-09-10T01:09:13

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

Explanation:

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

Answer 2 · 2016-09-10T01:15:25

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.
${log 4}^{3 x - 2} = log 1$ $log4^(3x-2) =log1$

By log properties, the exponent becomes a coefficient
( ${log x}^{a} = a log x$ $logx^a = alogx$ )
$(3 x - 2) log 4 = log 1$ $(3x-2)log4= log 1$

Divide both sides by log 4
$3 x - 2 = \frac{log 1}{log 4}$ $3x-2 = log1/log4$

$log 1 = 0$ $log1=0$ so $\frac{log 1}{log 4} = 0$ $log1/log4 = 0$

$3 x - 2 = 0$ $3x-2=0$

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

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

OR

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

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

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

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2 Answers

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