Solve for x 6e^(2x)+5e^x-4=0

Problem

6*e^(2*x)+5*e^x−4=0

Solution

1. Substitute a new variable to transform the equation into a quadratic form by letting u=e^x

6*u^2+5*u−4=0

2. Factor the quadratic equation by finding two numbers that multiply to 6⋅(−4)=−24 and add to 5 These numbers are 8 and −3

6*u^2+8*u−3*u−4=0

2*u*(3*u+4)−1*(3*u+4)=0

(2*u−1)*(3*u+4)=0

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

2*u−1=0⇒u=1/2

3*u+4=0⇒u=−4/3

4. Back-substitute e^x for u to solve for x

e^x=1/2

e^x=−4/3

5. Apply the natural logarithm to solve for x Note that e^x must be positive, so e^x=−4/3 has no real solution.

x=ln(1/2)

6. Simplify the expression using logarithm properties.

x=−ln(2)

Final Answer

x=−ln(2)

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