Solve (1/3)quad=27^3x+4

$$( \frac { 1 } { 3 } ) \quad = 27 ^ { 3 x + 4 }$$

Solve for x

$$x = -\frac{13}{9} = -1\frac{4}{9} \approx -1.444444444$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

$$27^{3x+4}=\frac{1}{3}$$

Take the logarithm of both sides of the equation.

$$\log(27^{3x+4})=\log(\frac{1}{3})$$

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

$$\left(3x+4\right)\log(27)=\log(\frac{1}{3})$$

Divide both sides by \log(27).

$$3x+4=\frac{\log(\frac{1}{3})}{\log(27)}$$

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

$$3x+4=\log_{27}\left(\frac{1}{3}\right)$$

Subtract 4 from both sides of the equation.

$$3x=-\frac{1}{3}-4$$

Divide both sides by 3.

$$x=-\frac{\frac{13}{3}}{3}$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{9\ln(3)}-\frac{13}{9}$$

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