Solve (5^{5}+0.01)^2-(5^5-0.01)^2=5^x26

$$( 5 ^ { 5 } + 0.01 ) ^ { 2 } - ( 5 ^ { 5 } - 0.01 ) ^ { 2 } = 5 ^ { x } 26$$

Solve for x

$$x=-\log_{5}\left(26\right)+3\approx 0.975630801$$

Steps Using Definition of Logarithm

Calculate 5 to the power of 5 and get 3125.

$$\left(3125+0.01\right)^{2}-\left(5^{5}-0.01\right)^{2}=5^{x}\times 26$$

Add 3125 and 0.01 to get 3125.01.

$$3125.01^{2}-\left(5^{5}-0.01\right)^{2}=5^{x}\times 26$$

Calculate 3125.01 to the power of 2 and get 9765687.5001.

$$9765687.5001-\left(5^{5}-0.01\right)^{2}=5^{x}\times 26$$

Calculate 5 to the power of 5 and get 3125.

$$9765687.5001-\left(3125-0.01\right)^{2}=5^{x}\times 26$$

Subtract 0.01 from 3125 to get 3124.99.

$$9765687.5001-3124.99^{2}=5^{x}\times 26$$

Calculate 3124.99 to the power of 2 and get 9765562.5001.

$$9765687.5001-9765562.5001=5^{x}\times 26$$

Subtract 9765562.5001 from 9765687.5001 to get 125.

$$125=5^{x}\times 26$$

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

$$5^{x}\times 26=125$$

Subtract 125 from both sides.

$$5^{x}\times 26-125=0$$

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

$$26\times 5^{x}-125=0$$

Add 125 to both sides of the equation.

$$26\times 5^{x}=125$$

Divide both sides by 26.

$$5^{x}=\frac{125}{26}$$

Take the logarithm of both sides of the equation.

$$\log(5^{x})=\log(\frac{125}{26})$$

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

$$x\log(5)=\log(\frac{125}{26})$$

Divide both sides by \log(5).

$$x=\frac{\log(\frac{125}{26})}{\log(5)}$$

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

$$x=\log_{5}\left(\frac{125}{26}\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(5)}-\log_{5}\left(26\right)+3$$

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