Solve 6^1+x+6^1+x=37

$$6 ^ { 1 + x } + 6 ^ { 1 + x } = 37$$

Solve for x

$$x=-\log_{6}\left(\frac{12}{37}\right)\approx 0.628438851$$

Steps Using Definition of Logarithm

Combine 6^{1+x} and 6^{1+x} to get 2\times 6^{1+x}.

$$2\times 6^{1+x}=37$$

Divide both sides by 2.

$$6^{1+x}=\frac{37}{2}$$

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

$$6^{x+1}=\frac{37}{2}$$

Take the logarithm of both sides of the equation.

$$\log(6^{x+1})=\log(\frac{37}{2})$$

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

$$\left(x+1\right)\log(6)=\log(\frac{37}{2})$$

Divide both sides by \log(6).

$$x+1=\frac{\log(\frac{37}{2})}{\log(6)}$$

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

$$x+1=\log_{6}\left(\frac{37}{2}\right)$$

Subtract 1 from both sides of the equation.

$$x=\frac{\ln(\frac{37}{2})}{\ln(6)}-1$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(6)}-\log_{6}\left(\frac{12}{37}\right)$$

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