Solve 5^x2-5x+6=1

$${ 5 }^{ x2-5x+6 } =1$$

Solve for x

$$x=\frac{x_{2}+6}{5}$$

Steps Using Definition of Logarithm

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

$$5^{-5x+x_{2}+6}=1$$

Take the logarithm of both sides of the equation.

$$\log(5^{-5x+x_{2}+6})=\log(1)$$

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

$$\left(-5x+x_{2}+6\right)\log(5)=\log(1)$$

Divide both sides by \log(5).

$$-5x+x_{2}+6=\frac{\log(1)}{\log(5)}$$

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

$$-5x+x_{2}+6=\log_{5}\left(1\right)$$

Subtract x_{2}+6 from both sides of the equation.

$$-5x=-\left(x_{2}+6\right)$$

Divide both sides by -5.

$$x=-\frac{x_{2}+6}{-5}$$

Solve for x_2

$$x_{2}=5x-6$$

Steps Using Definition of Logarithm

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

$$5^{x_{2}+6-5x}=1$$

Take the logarithm of both sides of the equation.

$$\log(5^{x_{2}+6-5x})=\log(1)$$

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

$$\left(x_{2}+6-5x\right)\log(5)=\log(1)$$

Divide both sides by \log(5).

$$x_{2}+6-5x=\frac{\log(1)}{\log(5)}$$

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

$$x_{2}+6-5x=\log_{5}\left(1\right)$$

Subtract -5x+6 from both sides of the equation.

$$x_{2}=-\left(6-5x\right)$$

Solve for x (complex solution)

$$x=-\frac{2\pi n_{1}i}{5\ln(5)}+\frac{x_{2}}{5}+\frac{6}{5}$$

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

Solve for x_2 (complex solution)

$$x_{2}=\frac{2\pi n_{1}i}{\ln(5)}+5x-6$$

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