How do you solve 10^(5x+2)=5^(4-x)?

Jul 10, 2016

$x = \frac{4 \ln \left(5\right) - 2 \ln \left(10\right)}{5 \ln \left(10\right) + \ln \left(5\right)} \approx 0.1397$

Explanation:

Using the property that $\ln \left({a}^{x}\right) = x \ln \left(a\right)$:

${10}^{5 x + 2} = {5}^{4 - x}$

$\implies \ln \left({10}^{5 x + 2}\right) = \ln \left({5}^{4 - x}\right)$

$\implies \left(5 x + 2\right) \ln \left(10\right) = \left(4 - x\right) \ln \left(5\right)$

$\implies 5 \ln \left(10\right) x + 2 \ln \left(10\right) = 4 \ln \left(5\right) - \ln \left(5\right) x$

$\implies 5 \ln \left(10\right) x + \ln \left(5\right) x = 4 \ln \left(5\right) - 2 \ln \left(10\right)$

$\implies \left(5 \ln \left(10\right) + \ln \left(5\right)\right) x = 4 \ln \left(5\right) - 2 \ln \left(10\right)$

$\implies x = \frac{4 \ln \left(5\right) - 2 \ln \left(10\right)}{5 \ln \left(10\right) + \ln \left(5\right)} \approx 0.1397$