205/VB02. problem / Math - Solved Math Problems

MR-205 / 1073. problem

Solve the following exponential equation:

$2^{x - 1} - 2^{x - 3} = 3^{x - 2} - 3^{x - 3}$

Source: Vene T. Bogoslavov: Collection of Solved Problems #2 - 1073..)

Algebra → Exponential equations and inequalities → Exponential equations

$2^{x - 1} - 2^{x - 3} = 3^{x - 2} - 3^{x - 3}$

2^{x - 1} - 2^{x - 3} = 3^{x - 2} - 3^{x - 3}

2^{x} \cdot 2^{- 1} - 2^{x} \cdot 2^{- 3} = 3^{x} \cdot 3^{- 2} - 3^{x} \cdot 3^{- 3}

\frac{2^{x}}{2} - \frac{2^{x}}{2^{3}} = \frac{3^{x}}{3^{2}} - \frac{3^{x}}{3^{3}}

2^{x} (\frac{1}{2} - \frac{1}{2^{3}}) = 3^{x} (\frac{1}{3^{2}} - \frac{1}{3^{3}})

2^{x} (\frac{2^{2} - 1}{2^{3}}) = 3^{x} (\frac{3 - 1}{3^{3}})

2^{x} (\frac{3}{2^{3}}) = 3^{x} (\frac{2}{3^{3}})

\frac{2^{x}}{3^{x}} = \frac{2 \cdot 2^{3}}{3^{3} \cdot 3}

\frac{2^{x}}{3^{x}} = \frac{2^{4}}{3^{4}}

{(\frac{2}{3})}^{x} = {(\frac{2}{3})}^{4}

$x = 4$