How do you solve $- 2 e^{7 v + 5} - 10 = - 17$ $-2e^(7v+5)-10=-17$ ?

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Answer 1 · 2018-05-14T12:34:54

Solution : $v \approx - 0.535$ $v ~~ -0.535$

$- 2 e^{7 v + 5} - 10 = - 17$ $-2 e^(7 v+5) -10 = -17$ or

$- 2 e^{7 v + 5} = 10 - 17$ $-2 e^(7 v+5) = 10-17$ or

$e^{7 v + 5} = \frac{7}{2}$ $e^(7 v+5) = 7/2$ Taking natural log on both sides we get,

$(7 v + 5) ln e = (ln 7 - ln 2); [ln e = 1]$ $(7 v+5) ln e = (ln 7 - ln 2) ; [ln e =1]$

$:. (7 v+5) ~~ 1.253 ;$

$:. 7 v ~~ 1.253-5 or 7 v ~~ -3.75 ;$

$:. v ~~ -3.75/7 ~~ -0.535 (3dp)$

Solution : $v ~~ -0.535$ [Ans]

How do you solve -2e^(7v+5)-10=-17−2e7v+5−10=−17-2e^(7v+5)-10=-17?