Notes: Exponential Equation with Base e — Solving e^{-x^2} = (e^x)^5 e^{-6}

Problem Setup

  • Given equation: e−x2=(ex)5⋅e−6.e^{-x^2} = \left(e^x\right)^5 \cdot e^{-6}.
  • Note: 1/e^6 = e^{-6}, so the right-hand side can be written using a single base e: (ex)5⋅e−6=e5x⋅e−6.\left(e^x\right)^5 \cdot e^{-6} = e^{5x} \cdot e^{-6}.
  • Therefore the equation becomes: e−x2=e5x−6.e^{-x^2} = e^{5x-6}.
  • Key principle: If two expressions have the same base and are equal, and the base is one-to-one (here base ee), then their exponents must be equal.

Step-by-Step Solution

  • Step 1: Apply exponent rules to rewrite the right-hand side:
    • (ex)5=e5x\left(e^x\right)^5 = e^{5x}
    • e−6=e−6e^{-6} = e^{-6}
    • So RHS = e5x⋅e−6=e5x−6.e^{5x} \cdot e^{-6} = e^{5x-6}.
  • Step 2: With bases equal, set exponents equal:
    • −x2=5x−6.-x^2 = 5x - 6.
  • Step 3: Move all terms to one side by adding x2x^2 to both sides (as in the transcript):
    • x2+5x−6=0.x^2 + 5x - 6 = 0.

Quadratic Factorization

  • Step 4: Factor the quadratic:
    • x2+5x−6=(x+6)(x−1).x^2 + 5x - 6 = (x+6)(x-1).
    • Verification: (x+6)(x−1)=x2−x+6x−6=x2+5x−6(x+6)(x-1) = x^2 - x + 6x - 6 = x^2 + 5x - 6.
  • Step 5: Apply the zero-product property:
    • If (x+6)(x−1)=0(x+6)(x-1) = 0, then either x=−6x = -6 or x=1.x = 1.

Solution Set

  • Final solution set: −6,  1.{ -6, \; 1 }.

Check / Verification (optional but recommended)

  • Check for x=−6x = -6:
    • LHS: e−(−6)2=e−36.e^{-(-6)^2} = e^{-36}.
    • RHS: (e−6)5⋅e−6=e−30⋅e−6=e−36.\left(e^{-6}\right)^5 \cdot e^{-6} = e^{-30} \cdot e^{-6} = e^{-36}.
    • Both sides match.
  • Check for x=1x = 1:
    • LHS: e−1.e^{-1}.
    • RHS: (e1)5⋅e−6=e5⋅e−6=e−1.\left(e^{1}\right)^5 \cdot e^{-6} = e^{5} \cdot e^{-6} = e^{-1}.
    • Both sides match.

Concepts and Rules Used

  • Exponent rules:
    • Power of a power: (ex)5=e5x\left(e^x\right)^5 = e^{5x}
    • Product of like bases: eA⋅eB=eA+Be^A \cdot e^B = e^{A+B}
  • One-to-one property of the exponential function: If eu=eve^u = e^v then u=vu = v.
  • Quadratic solving by factoring:
    • For a quadratic x2+5x−6=0x^2 + 5x - 6 = 0, factor to find roots x=−6,  1x = -6,\; 1.
  • Zero-product property: If ab=0ab = 0, then a=0a = 0 or b=0b = 0.
  • Domain note: Expressions involving ete^t are defined for all real tt, so both solutions are valid.

Connections to Related Ideas

  • This problem reinforces converting products of exponentials into a single exponential using base ee.
  • It also demonstrates why exponential equations with a common base can be solved by equating exponents, provided the base is positive and not equal to 1.
  • The approach mirrors typical steps in solving exponential equations: rewrite, equate exponents, solve resulting algebraic equation (often quadratic), and verify.