Solve (25/81)^2x+1=(729/125)

$$( \frac { 25 } { 81 } ) ^ { 2 x + 1 } = ( \frac { 729 } { 125 } )$$

Solve for x

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

Steps Using Definition of Logarithm

Use the rules of exponents and logarithms to solve the equation.

$$\left(\frac{25}{81}\right)^{2x+1}=\frac{729}{125}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{25}{81}\right)^{2x+1})=\log(\frac{729}{125})$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$\left(2x+1\right)\log(\frac{25}{81})=\log(\frac{729}{125})$$

Divide both sides by \log(\frac{25}{81}).

$$2x+1=\frac{\log(\frac{729}{125})}{\log(\frac{25}{81})}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$2x+1=\log_{\frac{25}{81}}\left(\frac{729}{125}\right)$$

Subtract 1 from both sides of the equation.

$$2x=-\frac{3}{2}-1$$

Divide both sides by 2.

$$x=-\frac{\frac{5}{2}}{2}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{2\ln(\frac{5}{9})}-\frac{\log_{\frac{5}{9}}\left(\frac{3125}{59049}\right)}{4}$$

$$n_{1}\in \mathrm{Z}$$