log1/7(x^2+x-5)=-1

Solve the equation log_{1/7}(x^2 + x - 5) = -1.

• Convert to an exponential form: x^2 + x - 5 = (1/7)^{-1} = 7.
• So x^2 + x - 12 = 0 → (x + 4)(x - 3) = 0 → x = -4 or x = 3.
• Check the domain: the argument must be positive. For x = -4 and x = 3, x^2 + x - 5 = 7 > 0, so both are valid.

Solutions: x = -4, 3.