Solve 29cosh x – 3cosh 2x = 38 for x, giving answers in terms of natural logarithms

Firstly, we need the equation to be in terms of one cosh argument if possible, which can be done by the identity cosh 2x = 2cosh²x - 1. Upon substitution and simplification we get 6cosh²x - 29cosh x + 35 = 0 which factorises to (3cosh x - 7)(2cosh x - 5) = 0. This gives two equations, namely, cosh x = 7/3, cosh x = 5/2.Next, the inverse cosh function can be determined by setting y = arcosh x => cosh y = x => (1/2)(e^y + e^-y) = x =>e^2y - 2xe^y + 1 = 0, which is a quadratic in e^y, which, the quadratic formula, solves to e^y = (1/2)(2x +- sqrt(4x² - 4)) = x +- sqrt(x² - 1). Taking natural logarithms then gives arcosh x = ln(x +- sqrt(x² - 1)). However, since cosh x >= 1 for all x, we take the positive square root, giving arcosh x = ln(x + sqrt(x² - 1))Finally, we can gain our final answers by using the log form of arcosh on 7/3 and 5/2 and simplifying to give x = ln(5/2 + sqrt(21)/2) & x = ln(7/3 + 2sqrt(10)/3).