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# C3 - OCR paul Sanders book - sample paper at the back watch

1. 6) two graphs of y = 3e^-x and y = 2e^3x-1

the graphs intersect at the point whose x is P

prove that P = 1/4ln(3e/2)

any ideas how to do this?
2. When the two graphs intersect

3e^{-x} = 2e^{3x} - 1

Solve for x.
3. lol i know that but than what do i do next
4. 3/2 e^(-x)= e^(3x-1)

Multiply through by e^x:
3/2=e^4x-1

Multiply through by e:
3e/2=e^4x

Ln both sides:
4x=Ln3e/2
x=(1/4)ln(3e/2)

5. Well i was trying to get you to think about that...
6. aaw thanks Ben i worked it out some longgg way using law of logs.
i ln -ed both sides

ln(2e^3x-1)=ln(3e^-x)
ln2 + 3x-1 = ln3 -x
4x = ln3 - ln2 + 1
4x = ln(3/2) + ln(e) note :- ln(e) =1
4x = ln(3e/2)
and than i finally got P = 1/4ln(3e/2)

yayy loll thanks BenSpurgen for your version though definately much easier. thought id post my way just in case anyone may be stuck on the same thing.
7. No problems

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