# Calculate: 4^{x^2-x+1}=8^x

## Expression: ${4}^{{x}^{2}-x+1}={8}^{x}$

Write the expression in exponential form with the base of $2$

${2}^{2{x}^{2}-2x+2}={8}^{x}$

Write the expression in exponential form with the base of $2$

${2}^{2{x}^{2}-2x+2}={2}^{3x}$

Since the bases are the same, set the exponents equal

$2{x}^{2}-2x+2=3x$

Move the variable to the left-hand side and change its sign

$2{x}^{2}-2x+2-3x=0$

Collect like terms

$2{x}^{2}-5x+2=0$

Write $-5x$ as a difference

$2{x}^{2}-x-4x+2=0$

Factor out $x$ from the expression

$x \times \left( 2x-1 \right)-4x+2=0$

Factor out $-2$ from the expression

$x \times \left( 2x-1 \right)-2\left( 2x-1 \right)=0$

Factor out $2x-1$ from the expression

$\left( 2x-1 \right) \times \left( x-2 \right)=0$

When the product of factors equals $0$, at least one factor is $0$

$\begin{array} { l }2x-1=0,\\x-2=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=\frac{ 1 }{ 2 },\\x-2=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=\frac{ 1 }{ 2 },\\x=2\end{array}$

The equation has $2$ solutions

\begin{align*}&\begin{array} { l }x_1=\frac{ 1 }{ 2 },& x_2=2\end{array} \\&\begin{array} { l }x_1=0.5,& x_2=2\end{array}\end{align*}

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