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8. (a) Simplify fully \[\frac{2x^2 + 9x - 5}{x^2 + 2x - 15}\] (3) Given that \[\ln(2x^2 + 9x - 5) = 1 + \ln(x^2 + 2x - 15), \ x \neq -5,\] (b) find x in terms of e - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 5
Question 1
![8.-(a)-Simplify-fully--\[\frac{2x^2-+-9x---5}{x^2-+-2x---15}\]--(3)--Given-that--\[\ln(2x^2-+-9x---5)-=-1-+-\ln(x^2-+-2x---15),-\-x-\neq--5,\]--(b)-find-x-in-terms-of-e-Edexcel-A-Level Maths Pure-Question 1-2010-Paper 5.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/MathsPure_2010_5_1_QP_8_2010 .jpg)
8. (a) Simplify fully \[\frac{2x^2 + 9x - 5}{x^2 + 2x - 15}\] (3) Given that \[\ln(2x^2 + 9x - 5) = 1 + \ln(x^2 + 2x - 15), \ x \neq -5,\] (b) find x in terms o... show full transcript
Worked Solution & Example Answer:8. (a) Simplify fully \[\frac{2x^2 + 9x - 5}{x^2 + 2x - 15}\] (3) Given that \[\ln(2x^2 + 9x - 5) = 1 + \ln(x^2 + 2x - 15), \ x \neq -5,\] (b) find x in terms of e - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 5
Step 1
Simplify fully
Answer
To simplify the fraction ( \frac{2x^2 + 9x - 5}{x^2 + 2x - 15} ), we first factor the numerator and the denominator.
Numerator: We look for two numbers that multiply to (-10) (which is (2 \times -5)) and add to (9). These numbers are (10) and (-1):
[ 2x^2 + 10x - x - 5 = 2x(x + 5) - 1(x + 5) = (2x - 1)(x + 5) ]
Denominator: We need to factor ( x^2 + 2x - 15 ). The numbers that multiply to (-15) and add to (2) are (5) and (-3):
[ x^2 + 5x - 3x - 15 = (x + 5)(x - 3) ]
Now we rewrite the expression:
[ \frac{(2x - 1)(x + 5)}{(x + 5)(x - 3)} ]
Cancelling ( (x + 5) ) from the numerator and denominator (noting that ( x \neq -5 )), we get:
[ \frac{2x - 1}{x - 3} ]
Step 2
find x in terms of e
Answer
Given the equation:
[ \ln(2x^2 + 9x - 5) = 1 + \ln(x^2 + 2x - 15) ]
We can rearrange this to:
[ \ln(2x^2 + 9x - 5) - \ln(x^2 + 2x - 15) = 1 ]
Using the properties of logarithms, this simplifies to:
[ \ln\left( \frac{2x^2 + 9x - 5}{x^2 + 2x - 15} \right) = 1 ]
Exponentiating both sides gives us:
[ \frac{2x^2 + 9x - 5}{x^2 + 2x - 15} = e ]
We already simplified the left side as ( \frac{2x - 1}{x - 3} ). Setting it equal to e:
[ \frac{2x - 1}{x - 3} = e ]
Cross multiplying yields:
[ 2x - 1 = e(x - 3) ]
Simplifying further:
[ 2x - 1 = ex - 3e ]
Rearranging terms gives:
[ 2x - ex = 1 - 3e ]
Factoring out (x):
[ x(2 - e) = 1 - 3e ]
Finally, solving for (x):
[ x = \frac{1 - 3e}{2 - e} ]