Exponential and Logarithmic Equations

Question

e^x + 45e^−x − 14 = 0

solve for x, round to four decimal places.

Answer 1

Hello Caitlin,

You can turn this equation into an equation that is quadratic in form by multiplying both sides by e^x. That is,

e^x + 45e^-x - 14 = 0

e^x(e^x + 45e^-x - 14) = e^x⋅0

Distribute and add exponents in the above equation to get

e^2x + 45e⁰ - 14e^x = 0

e^2x + 45 - 14e^x = 0 (using the fact that e⁰ = 1.)

e^2x - 14e^x + 45 = 0.

Now substitute u = e^x. Then, since u² = (e^x)² = e^2x, the above equation can be rewritten as

u² - 14u + 45 = 0.

(Since the original equation can be turned into a quadratic equation by using an appropriate substitution, we say that the original equation is quadratic in form.) Now it is easy to solve the last equation by factoring.

(u - 9)(u - 5) = 0

u - 9 = 0 or u - 5 = 0

u = 9 or u = 5.

To complete the solution, we substitute e^x back in for u, and then take the natural log of each side.

e^x = 9 or e^x = 5.

Ln(e^x) = Ln(9) or Ln(e^x) = Ln(5)

x = Ln(9) or x = Ln(5)

x ≅ 2.1972 or x ≅ 1.6094 (rounded to 4 decimal places.)

Hope that helps! Let me know if you need any further help.

William

Answer 2

e^x + 45e^(-x) - 14 = 0 <--- given

(e^x)*(e^x) + 45*e^-x * e^x - 14*e^x = 0 <--- multiplies everything by e^x

(e^x)^2 + 45 - 14*e^x = 0 <--- adds the exponents for the same base e; note that -x+x = 0 and e^0 = 1

(e^x)^2 - 14*e^x + 45 = 0

M^2 - 14M + 45 = 0 <--- substitution M=e^x

(M - 9 )(M - 5 ) = 0

M-9 = 0 or M-5 = 0

M=9 or M=5

e^x=9 or e^x = 5

x = ln9 or x = ln5

Answer 3

Multiply through by ex to get e^2x -14e^x+45 = 0 Treat e^x as the variable in a quadratic equation to get

e^x = {-(-14) +- SQRT( (-14)2 -4(1)(45) )}/2 to get e^x = 7 +- 2 or e^x = 5 and 9

Therefore; x = ln(5) = 1.6094 and x = ln(9) = 2.1972