Solve 18*2^x=2

$$18 \cdot 2 ^ { x } = 2$$

Solve for x

$$x=-2\log_{2}\left(3\right)\approx -3.169925001$$

Steps Using Definition of Logarithm

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

$$18\times 2^{x}=2$$

Divide both sides by 18.

$$2^{x}=\frac{1}{9}$$

Take the logarithm of both sides of the equation.

$$\log(2^{x})=\log(\frac{1}{9})$$

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

$$x\log(2)=\log(\frac{1}{9})$$

Divide both sides by \log(2).

$$x=\frac{\log(\frac{1}{9})}{\log(2)}$$

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

$$x=\log_{2}\left(\frac{1}{9}\right)$$

Solve for x (complex solution)

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

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