Solve for x e^x-30e^(-x)-1=0

Problem

e^x−30*e^(−x)−1=0

Solution

1. Multiply the entire equation by e^x to eliminate the negative exponent.

e^x*(e^x−30*e^(−x)−1)=0⋅e^x

(e^x)^2−30−e^x=0

2. Rearrange the terms into a standard quadratic form.

(e^x)^2−e^x−30=0

3. Substitute a new variable u=e^x to simplify the equation.

u^2−u−30=0

4. Factor the quadratic equation by finding two numbers that multiply to −30 and add to −1

(u−6)*(u+5)=0

5. Solve for u by setting each factor to zero.

u=6

u=−5

6. Substitute back e^x for u and solve for x

e^x=6⇒x=ln(6)

e^x=−5

7. Identify extraneous solutions. Since e^x must always be greater than zero for real x the equation e^x=−5 has no real solution.

x=ln(6)

Final Answer

e^x−30*e^(−x)−1=0⇒x=ln(6)

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