Solve 4*6^x+1=10/9?

$$4 \cdot 6 ^ { x } + 1 = \frac { 10 } { 9 } ?$$

Solve for x

$$x=-2$$

Steps Using Definition of Logarithm

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

$$4\times 6^{x}+1=\frac{10}{9}$$

Subtract 1 from both sides of the equation.

$$4\times 6^{x}=\frac{1}{9}$$

Divide both sides by 4.

$$6^{x}=\frac{1}{36}$$

Take the logarithm of both sides of the equation.

$$\log(6^{x})=\log(\frac{1}{36})$$

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

$$x\log(6)=\log(\frac{1}{36})$$

Divide both sides by \log(6).

$$x=\frac{\log(\frac{1}{36})}{\log(6)}$$

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

$$x=\log_{6}\left(\frac{1}{36}\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(6)}-2$$

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