Solve 125*25^2r+1=1

$$125 \cdot 25 ^ { 2 r + 1 } = 1$$

Solve for r

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

Steps Using Definition of Logarithm

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

$$125\times 25^{2r+1}=1$$

Divide both sides by 125.

$$25^{2r+1}=\frac{1}{125}$$

Take the logarithm of both sides of the equation.

$$\log(25^{2r+1})=\log(\frac{1}{125})$$

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

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

Divide both sides by \log(25).

$$2r+1=\frac{\log(\frac{1}{125})}{\log(25)}$$

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

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

Subtract 1 from both sides of the equation.

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

Divide both sides by 2.

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