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Solve exponential equation of the form \\[ a^{kx}=b^{kx+m}. \\]

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Solve the equation \\[ \\var{base1}^{\\simplify{{k} x}} = \\var{base2}^{\\simplify{{k}x+{m}}} \\]

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Since some powers are the same, begin by rewriting the equations as $\\var{base1}^{\\var{k}x} = \\var{base2}^{\\var{k}x}\\times  \\var{base2}^{\\var{m}}$.

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Then group the terms with the same power to get: $\\left(\\frac{\\var{base1}}{\\var{base2}}\\right)^{\\var{k}x} =\\var{base2}^{\\var{m}}$.

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Take logs of both sides to obtain: $ \\var{k}x = \\frac{\\ln(\\var{base2}^{\\var{m}})}{\\ln\\left(\\var{base1}/\\var{base2}\\right)}$.

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Therefore, $x=\\frac{\\ln(\\var{base2}^{\\var{m}})}{\\var{k}\\ln\\left(\\var{base1}/\\var{base2}\\right)}=${ (1/{k})*ln({base2}^{m})/ln({base1}/{base2}) }

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$x=$[[0]]

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Start by rewriting the equations as $\\var{base1}^{\\var{k}x} = \\var{base2}^{\\var{k}x}\\times  \\var{base2}^{\\var{m}}$.

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