## Algebra 2 (1st Edition)

$$x=\frac{31}{15}$$
To solve this problem, since there are only logarithms with the same base on either side of the equation, we cancel out the logarithms. Then, we solve for x. $$18x+7=3x+38 \\ x=\frac{31}{15}$$ Checking for extraneous solutions, by plugging the value of x back into the equation and checking if it yields a true statement, we see that the solution is correct.