CALCULUS!! HELP!?!?!?!!!?

Question

how do i algebraically solve

e^x + e^-x = 10

thanks!!

Answer 1

e^x - e^-x = 2cosh(x) , so that

2cosh(x) = 10

cosh(x) = 5

x = cosh^-1(5)

x = 2.2924316

Answer 2

Let y =e^x

e^x + e^-x = 10

y + 1/y = 10

y^2-10y+1=0

Solve using quadratic eqution.

Resubstitute for x

Solve using logarithms.