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The number of real solutions of the equation e4x + 4e3x – 58e2x + 4ex + 1 = 0, is……………

0 2 Views | Posted 2 weeks ago
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  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    2 weeks ago

    e 4 x + 4 e 3 x − 5 8 e 2 x + 4 e x + 1 = 0

    ( e 2 x + 1 e 2 x ) + 4 ( e x + 1 e x ) − 5 8 = 0

    ⇒ ( e x + 1 e x + 2 ) 2 = 6 4

    e x = 6 ± 3 2 2 = 3 ± 2 2

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