Solve x=5^5x5^-y

$$x= { 5 }^{ 5 } x { 5 }^{ -y }$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&y=5\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 5 to the power of 5 and get 3125.

$$x=3125x\times 5^{-y}$$

Subtract 3125x\times 5^{-y} from both sides.

$$x-3125x\times 5^{-y}=0$$

Combine all terms containing x.

$$\left(1-3125\times 5^{-y}\right)x=0$$

The equation is in standard form.

$$\left(-\frac{3125}{5^{y}}+1\right)x=0$$

Divide 0 by 1-3125\times 5^{-y}.

$$x=0$$

Solve for y

$$\left\{\begin{matrix}\\y=5\text{, }&\text{unconditionally}\\y\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps Using Definition of Logarithm

Calculate 5 to the power of 5 and get 3125.

$$x=3125x\times 5^{-y}$$

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

$$3125x\times 5^{-y}=x$$

Divide both sides by 3125x.

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

Take the logarithm of both sides of the equation.

$$\log(5^{-y})=\log(\frac{1}{3125})$$

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

$$-y\log(5)=\log(\frac{1}{3125})$$

Divide both sides by \log(5).

$$-y=\frac{\log(\frac{1}{3125})}{\log(5)}$$

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

$$-y=\log_{5}\left(\frac{1}{3125}\right)$$

Divide both sides by -1.

$$y=-\frac{5}{-1}$$