How to solve 2 ^ (3x) = 9

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 2 ^ (3x) = 9 | Equatix

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$$2^{3x}=9$$

$x = \frac{2 \log_{2} {(3)}}{3} \approx 1.056641667$

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

$$2^{3x}=9$$

Take the logarithm of both sides of the equation.

$$\log(2^{3x})=\log(9)$$

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

$$3x\log(2)=\log(9)$$

Divide both sides by $\log(2)$.

$$3x=\frac{\log(9)}{\log(2)}$$

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

$$3x=\log_{2}\left(9\right)$$

Divide both sides by $3$.

$$x=\frac{2\log_{2}\left(3\right)}{3}$$