solve each of the following equations.㏒₇(9x+38) - ㏒₇(x+2) = ㏒₉81

first you need to determine the defined range
[tex] log_{7} [/tex](9x + 38) - [tex] log_{7} [/tex](x + 2) = [tex] log_{9} [/tex](81), x∈ (-2, + ∞)using the [tex] log_{a} [/tex](x) - [tex] log_{a} [/tex](y) = [tex] log_{a} [/tex] ([tex] \frac{x}{y} [/tex]), simplify the expression
[tex] log_{7} [/tex] ([tex] \frac{9x+38}{x+2} [/tex]) =  [tex] log_{9} [/tex](81)
write the number in the second set of parenthesis in exponential form
[tex] log_{7} [/tex] ([tex] \frac{9x+38}{x+2} [/tex]) =  [tex] log_{9} [/tex] (9²)
using [tex] log_{a} [/tex] ([tex] a^{x} [/tex]) = x, simplify the expression
[tex] log_{7} [/tex] ([tex] \frac{9x+38}{x+2} [/tex]) = 2
the expression [tex] log_{a} [/tex](x) = b is equal to x = [tex] a^{b} [/tex]
[tex] \frac{9x+38}{x+2} [/tex] = 7²
evaluate the power
[tex] \frac{9x+38}{x+2} [/tex] = 49
multiply both sides of the equation by x + 2
9x + 38 = 49(x + 2)
distribute 49 through the parenthesis
9x + 38 = 49x + 98
move the variable to the left side and change its sign
9x - 49x + 38 = 98
move the constant to the right side and change its sign
9x - 49x = 98 - 39
collect the like terms
-40x = 98 - 38
subtract the numbers
-40x = 60
divide both sides of the equation by -40
x = -[tex] \frac{3}{2} [/tex], x ∈ (-2, + ∞)
finally,, check if the solution is in the defined range
x = -[tex] \frac{3}{2} [/tex]
this means that the correct answer to your question is x = -[tex] \frac{3}{2} [/tex]
let me know if you have any further questions
:)