Solve for x (e^x+e^(-x))/2=15

Problem

(e^x+e^(−x))/2=15

Solution

1. Multiply both sides of the equation by 2 to clear the fraction.

(e^x+e^(−x))/2=15

e^x+e^(−x)=30

2. Eliminate the negative exponent by multiplying every term in the equation by e^x

e^x*(e^x+e^(−x))=30*(e^x)

(e^x)^2+1=30*e^x

3. Rearrange the equation into a quadratic form by moving all terms to one side.

(e^x)^2−30*e^x+1=0

4. Substitute a variable u=e^x to make the quadratic structure clear.

u^2−30*u+1=0

5. Apply the quadratic formula u=(−b±√(,b^2−4*a*c))/(2*a) where a=1 b=−30 and c=1

u=(30±√(,(−30)^2−4*(1)*(1)))/(2*(1))

u=(30±√(,900−4))/2

u=(30±√(,896))/2

6. Simplify the radical √(,896)=√(,64⋅14)=8√(,14)

u=(30±8√(,14))/2

u=15±4√(,14)

7. Solve for x by substituting e^x back in for u and taking the natural logarithm of both sides.

e^x=15±4√(,14)

x=ln(15±4√(,14))

Final Answer

x=ln(15±4√(,14))

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