How to solve 3 ^ x = 729

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 3 ^ x = 729 | Equatix

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$$3^{x}=729$$

$x=6$

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

$$3^{x}=729$$

Take the logarithm of both sides of the equation.

$$\log(3^{x})=\log(729)$$

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

$$x\log(3)=\log(729)$$

Divide both sides by $\log(3)$.

$$x=\frac{\log(729)}{\log(3)}$$

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

$$x=\log_{3}\left(729\right)$$

$x=\frac{2\pi n_{1}i}{\ln(3)}+6$
$n_{1}\in \mathrm{Z}$