Solve for x 7^(2x)+7^x-6=0

Problem

7^(2*x)+7^x−6=0

Solution

1. Substitute a new variable to transform the equation into a quadratic form. Let u=7^x

2. Rewrite the equation using the substitution, noting that 7^(2*x)=(7^x)^2=u^2

u^2+u−6=0

3. Factor the quadratic equation by finding two numbers that multiply to −6 and add to 1

(u+3)*(u−2)=0

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

u=−3

u=2

5. Back-substitute u=7^x to solve for x

7^x=−3

7^x=2

6. Evaluate the solutions. Since 7^x must be positive for all real x the equation 7^x=−3 has no real solution.

7. Apply logarithms to solve the remaining equation 7^x=2

ln(7^x)=ln(2)

x*ln(7)=ln(2)

x=ln(2)/ln(7)

Final Answer

x=ln(2)/ln(7)

Want more problems? Check here!