# How do you solve this logarithmic function?

## ${\left(\frac{16}{25}\right)}^{x + 3} = {\left(\frac{125}{64}\right)}^{x - 1}$

Nov 17, 2016

#### Explanation:

Use either the natural or base 10 logarithm on both sides. (I will use the natural logarithm):

$\ln \left({\left(\frac{16}{25}\right)}^{x + 3}\right) = \ln \left({\left(\frac{125}{64}\right)}^{x - 1}\right)$

Use the property of all logarithms ${\log}_{b} \left({a}^{c}\right) = \left(c\right) {\log}_{b} \left(a\right)$:

$\left(x + 3\right) \ln \left(\frac{16}{25}\right) = \left(x - 1\right) \ln \left(\frac{125}{64}\right)$

Use the distributive property on both sides:

$\left(x\right) \ln \left(\frac{16}{25}\right) + \left(3\right) \ln \left(\frac{16}{25}\right) = \left(x\right) \ln \left(\frac{125}{64}\right) - \ln \left(\frac{125}{64}\right)$

Move the x terms to the left and the constant terms to the right:

$\left(x\right) \ln \left(\frac{16}{25}\right) - \left(x\right) \ln \left(\frac{125}{64}\right) = - \left(3\right) \ln \left(\frac{16}{25}\right) - \ln \left(\frac{125}{64}\right)$

Factor out x on the left:

$\left(x\right) \left(\ln \left(\frac{16}{25}\right) - \ln \left(\frac{125}{64}\right)\right) = - \left(3\right) \ln \left(\frac{16}{25}\right) - \ln \left(\frac{125}{64}\right)$

Divide both sides by the coefficient of x:

$x = - \frac{\left(3\right) \ln \left(\frac{16}{25}\right) + \ln \left(\frac{125}{64}\right)}{\ln \left(\frac{16}{25}\right) - \ln \left(\frac{125}{64}\right)}$

Multiply the denominator by the -1 in front:

$x = \frac{\left(3\right) \ln \left(\frac{16}{25}\right) + \ln \left(\frac{125}{64}\right)}{\ln \left(\frac{125}{64}\right) - \ln \left(\frac{16}{25}\right)}$

$x = - 0.6$