Solve 3^x+118/3=19

$${ 3 }^{ x+1 } \frac{ 18 }{ 3 } =19$$

Solve for x

$$x=-\log_{3}\left(\frac{2}{19}\right)-2\approx 0.049214106$$

Steps Using Definition of Logarithm

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

$$6\times 3^{x+1}=19$$

Divide both sides by 6.

$$3^{x+1}=\frac{19}{6}$$

Take the logarithm of both sides of the equation.

$$\log(3^{x+1})=\log(\frac{19}{6})$$

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

$$\left(x+1\right)\log(3)=\log(\frac{19}{6})$$

Divide both sides by \log(3).

$$x+1=\frac{\log(\frac{19}{6})}{\log(3)}$$

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

$$x+1=\log_{3}\left(\frac{19}{6}\right)$$

Subtract 1 from both sides of the equation.

$$x=\frac{\ln(\frac{19}{6})}{\ln(3)}-1$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(3)}-\log_{3}\left(\frac{2}{19}\right)-2$$

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