Ask Question
16 June, 01:47

I hate to ask but i am stumped, how do you work out 2e^5x - 7e^2x - 15e^-x=0

+3
Answers (1)
  1. 16 June, 02:56
    0
    Use some variable to take the place of e^x temporarily. Say

    u=e^x

    =>2u^5-7u^2-15/u=0

    => 2u^6-7u^3-15=0 and u cannot be equal to 0.

    use another intermediate variable: m=u^3 (m=e^ (3x))

    =>2m^2-7m-15=0

    => (m-5) * (2m+3) = 0

    => m=5 and m=-3/2

    => e^3x=5 and e^3x=-3/2

    => 3x=Ln (5) and 3x=Ln (-3/2) (Ln (-3/2) does not exist so this solution is extraneous)

    => x=Ln (5) / 3
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “I hate to ask but i am stumped, how do you work out 2e^5x - 7e^2x - 15e^-x=0 ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers