Solve -8*4^x+1+18*4^x+1=10*4^-2

$$- 8 \times 4 ^ { x + 1 } + 18 \cdot 4 ^ { x + 1 } = 10 \times 4 ^ { - 2 }$$

Solve for x

$$x=-3$$

Steps Using Definition of Logarithm

Combine -8\times 4^{x+1} and 18\times 4^{x+1} to get 10\times 4^{x+1}.

$$10\times 4^{x+1}=10\times 4^{-2}$$

Calculate 4 to the power of -2 and get \frac{1}{16}.

$$10\times 4^{x+1}=10\times \frac{1}{16}$$

Multiply 10 and \frac{1}{16} to get \frac{5}{8}.

$$10\times 4^{x+1}=\frac{5}{8}$$

Divide both sides by 10.

$$4^{x+1}=\frac{\frac{5}{8}}{10}$$

Express \frac{\frac{5}{8}}{10} as a single fraction.

$$4^{x+1}=\frac{5}{8\times 10}$$

Multiply 8 and 10 to get 80.

$$4^{x+1}=\frac{5}{80}$$

Reduce the fraction \frac{5}{80} to lowest terms by extracting and canceling out 5.

$$4^{x+1}=\frac{1}{16}$$

Take the logarithm of both sides of the equation.

$$\log(4^{x+1})=\log(\frac{1}{16})$$

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

$$\left(x+1\right)\log(4)=\log(\frac{1}{16})$$

Divide both sides by \log(4).

$$x+1=\frac{\log(\frac{1}{16})}{\log(4)}$$

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

$$x+1=\log_{4}\left(\frac{1}{16}\right)$$

Subtract 1 from both sides of the equation.

$$x=-2-1$$

Solve for x (complex solution)

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

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