ln(2x^2 + 9x – 5) = 1 + ln(x^2 + 2x – 15). Express x in terms of e

ln(2x2+ 9x – 5) = 1 + ln(x2 + 2x – 15)ln(2x2+ 9x – 5) – ln(x2 + 2x – 15)= 1 ln((2x2+ 9x – 5)/(x2 + 2x – 15)) = 1(2x2+ 9x – 5)/(x2 + 2x – 15) = e(2x – 1)(x + 5)/(x + 5)(x – 3) = e (2x – 1)/(x – 3) = e2x – 1 = e(x – 3)2x – 1 = ex – 3e2x – ex = 1 – 3ex(2 – e) = 1 – 3ex = (1 – 3e)/(2 – e)

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Answered by Sylvain F. Maths tutor

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