Solve (-1/5)^5*(1/5)^7div5^-1=-(5)^3y-1

$$( - \frac { 1 } { 5 } ) ^ { 5 } \times ( \frac { 1 } { 5 } ) ^ { 7 } \div 5 ^ { - 1 } = - ( 5 ) ^ { 3 y - 1 }$$

Solve for y

$$y = -\frac{10}{3} = -3\frac{1}{3} \approx -3.333333333$$

Steps Using Definition of Logarithm

Calculate -\frac{1}{5} to the power of 5 and get -\frac{1}{3125}.

$$\frac{-\frac{1}{3125}\times \left(\frac{1}{5}\right)^{7}}{5^{-1}}=-5^{3y-1}$$

Calculate \frac{1}{5} to the power of 7 and get \frac{1}{78125}.

$$\frac{-\frac{1}{3125}\times \frac{1}{78125}}{5^{-1}}=-5^{3y-1}$$

Multiply -\frac{1}{3125} and \frac{1}{78125} to get -\frac{1}{244140625}.

$$\frac{-\frac{1}{244140625}}{5^{-1}}=-5^{3y-1}$$

Calculate 5 to the power of -1 and get \frac{1}{5}.

$$\frac{-\frac{1}{244140625}}{\frac{1}{5}}=-5^{3y-1}$$

Divide -\frac{1}{244140625} by \frac{1}{5} by multiplying -\frac{1}{244140625} by the reciprocal of \frac{1}{5}.

$$-\frac{1}{244140625}\times 5=-5^{3y-1}$$

Multiply -\frac{1}{244140625} and 5 to get -\frac{1}{48828125}.

$$-\frac{1}{48828125}=-5^{3y-1}$$

Swap sides so that all variable terms are on the left hand side.

$$-5^{3y-1}=-\frac{1}{48828125}$$

Divide both sides by -1.

$$5^{3y-1}=\frac{-\frac{1}{48828125}}{-1}$$

Express \frac{-\frac{1}{48828125}}{-1} as a single fraction.

$$5^{3y-1}=\frac{-1}{48828125\left(-1\right)}$$

Cancel out -1 in both numerator and denominator.

$$5^{3y-1}=\frac{1}{48828125}$$

Take the logarithm of both sides of the equation.

$$\log(5^{3y-1})=\log(\frac{1}{48828125})$$

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

$$\left(3y-1\right)\log(5)=\log(\frac{1}{48828125})$$

Divide both sides by \log(5).

$$3y-1=\frac{\log(\frac{1}{48828125})}{\log(5)}$$

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

$$3y-1=\log_{5}\left(\frac{1}{48828125}\right)$$

Add 1 to both sides of the equation.

$$3y=-11-\left(-1\right)$$

Divide both sides by 3.

$$y=-\frac{10}{3}$$

Solve for y (complex solution)

$$y=\frac{2\pi n_{1}i}{3\ln(5)}-\frac{10}{3}$$

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