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P6 Ex 1B qs 5

Given that x is so small that terms in x^3 and higher powers of may be disregarded, show that

ln [ (1+2x)^2 / 1-3x ]= 7x+1/2x^2

im getting x+1/2x^2... but i dont c any wrong that i do. please help.
Reply 1
Avatar for Fermat
Fermat
8
ln [ (1+2x)^2 / 1-3x ] = 2ln(1 + 2x) - ln(1 - 3x)
= 2(2x - 4x²/2 + ...) - (-3x - 9x²/2 - ...)
= 4x - 4x² + 3x + 4.5x²
= 7x + 0.5x²
=========
Reply 2
Avatar for Fermat
Fermat
8
I think you subtracted the 3x instead of adding it :smile:
Reply 3
Avatar for El Stevo
El Stevo
11
i assume you realised...

ln[{(1+2x)^2}/(1-3x)] = 2ln(1+2x) - ln(1-3x)

and using...

ln(1+x) = x - (x^2)/2 (as ignoring terms in (x^3) or greater...)

ln(1+2x) = 2x - {(2x)^2}/2
ln(1+2x) = 2x - 4(x^2)/2
ln(1+2x) = 2x - 2(x^2)

ln(1-3x) = -3x - {(-3x)^2}/2
ln(1-3x) = -3x - 9(x^2)/2
ln(1-3x) = -3x - (9/2)(x^2)

ln[{(1+2x)^2}/(1-3x)] = 2ln(1+2x) - ln(1-3x)
ln[{(1+2x)^2}/(1-3x)] = 2(2x - 2(x^2)) - (-3x - (9/2)(x^2))
ln[{(1+2x)^2}/(1-3x)] = 4x - 4(x^2)) + 3x + (9/2)(x^2))
ln[{(1+2x)^2}/(1-3x)] = 7x + (1/2)(x^2)

um: did the rhs of your equation have any brackets we should know about?

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