Solve 3^alpha-5*3+6=0

$$3 ^ { \alpha } - 5 \times 3 + 6 = 0$$

Solve for α

$$\alpha =2$$

Steps Using Definition of Logarithm

Reorder the terms.

$$-5\times 3+3^{\alpha }+6=0$$

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

$$3^{\alpha }-9=0$$

Add 9 to both sides of the equation.

$$3^{\alpha }=9$$

Take the logarithm of both sides of the equation.

$$\log(3^{\alpha })=\log(9)$$

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

$$\alpha \log(3)=\log(9)$$

Divide both sides by \log(3).

$$\alpha =\frac{\log(9)}{\log(3)}$$

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

$$\alpha =\log_{3}\left(9\right)$$