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Question

# Solve the equation.$3^{2 x-1}=6^{x} 3^{1-x}$

Solution

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Is given the equation

\begin{aligned} 3 ^ { 2 x - 1 } = 6 ^ { x } \cdot 3 ^ { 1 - x } \end{aligned}

Since $6=2\cdot 3$, $6 ^ { x } = ( 2 \cdot 3 ) ^ { x } = 2 ^ { x } - 3 ^ { x }$ (product rule for exponential) then the equation can be written as

\begin{aligned} 3 ^ { 2 x - 1 } = 2 ^ { x } \cdot 3 ^ { x } \cdot 3 ^ { 1 - x } \equiv \frac { 3 ^ { 2 x - 1 } } { 3 ^ { x } \cdot 3 ^ { 1 - x } } = 2 ^ { x } \end{aligned}

Using the product and quotient rule for exponential in the left side of the equation, we have that

\begin{aligned} 3 ^ { 2 x - 1 - x - ( 1 - x ) } = 2 ^ { x } \Rightarrow 3 ^ { 2 x - 2 } = 2 ^ { x }\\ \Rightarrow \frac { 3 ^ { 2 x } } { 9 } = 2 ^ { x } \Rightarrow \frac { 9 ^ { x } } { 9 } = 2 ^ { x } \Rightarrow \frac { 9 ^ { x } } { 2 ^ { x } } = 9\\ \Rightarrow ( 9 / 2 ) ^ { x } = 9 \end{aligned}

where we have used the product and quotient rules and the power rule to write $3 ^ { 2 x } = \left( 3 ^ { 2 } \right) ^ { x } = 9 ^ { x }$. Using now the equality rule for logarithm choosing as base $b = 9 / 2$ result

\begin{aligned} \log _ { 9 / 2 } ( 9 / 2 ) ^ { x } = \log _ { 9 / 2 } 9 \end{aligned}

Now, if we use the power rule its result

\begin{aligned} x = \log _ { 9 / 2 } 9 \end{aligned}

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