[FP2] Solve: 3 cosh x - 4 sinh x = 7

First write cosh and sinh as exponentials, from their respective definitions:sinh x = 1/2 (e^x  - e^-x)cosh x = 1/2 (e^x  + e^-x​)So: 3 * 1/2 (e^x  + e^-x​) - 4 * 1/2 (e^x  - e^-x) = 7Multiply by 2: 3 * (e^x  + e^-x​) - 4 * (e^x  - e^-x) = 14Expanding gives: 3 * e^x  + 3 * e^-x  - 4 * e^x + 4 * e^-x = 14Collecting terms: -e^x + 7 * e^-x = 14So:  e^x - 7 * e^-x + 14 = 0Since e^x is not 0, we can multiply by it:e^x * e^x - 7 * e^-x *e^x + 14 * e^x = 0e^2x - 7 + 14 * e^x = 0This is a quadratic in e^x, and we can use the quadratic formula to obtain:e^x = -7 +- 2 root 14But e^x > 0, so we discount the negative root:e^x = -7 + 2 root 14x = ln ( -7 + 2 root 14)Which can easily be checked with a calculator.