Solving Exponential Equations Math Example 4

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Example 4

hard
Solve e2x−5ex+6=0e^{2x} - 5e^x + 6 = 0.

Solution

  1. 1
    Let u=exu = e^x, so u2−5u+6=0u^2 - 5u + 6 = 0. Factor: (u−2)(u−3)=0(u-2)(u-3) = 0, giving u=2u = 2 or u=3u = 3.
  2. 2
    Back-substitute: ex=2⇒x=ln⁡2e^x = 2 \Rightarrow x = \ln 2, and ex=3⇒x=ln⁡3e^x = 3 \Rightarrow x = \ln 3.

Answer

x=ln⁡2 or x=ln⁡3x = \ln 2 \text{ or } x = \ln 3
Exponential equations that are quadratic in form can be solved by substitution. Letting u=exu = e^x converts the equation to a polynomial. After solving for uu, reject any negative solutions since ex>0e^x > 0 for all xx. Here both solutions are positive, so both are valid.

About Solving Exponential Equations

Solving exponential equations means finding the unknown variable trapped in an exponent by applying logarithms to both sides, using the power rule to bring the exponent down, and then isolating the variable with standard algebra.

Learn more about Solving Exponential Equations →

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Example 1 easy

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Example 3 medium

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