8. (a) Simplify fully $$\frac{2x^2 + 9x - 5}{x^2 + 2x - 15}$$ (3) Given that $$\ln(2x^2 + 9x - 5) = 1 + \ln(x^2 + 2x - 15), \quad x \neq -5,$$ (b) find x in terms of e - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 2

To solve for x, we start with the given equation:

$\ln(2x^2 + 9x - 5) = 1 + \ln(x^2 + 2x - 15).$

1. Isolate the logarithmic expressions:

   • Rewrite the equation as: $\ln(2x^2 + 9x - 5) - \ln(x^2 + 2x - 15) = 1.$
     This simplifies further to: $\ln\left(\frac{2x^2 + 9x - 5}{x^2 + 2x - 15}\right) = 1.$

2. Exponentiate both sides:

   • Converting from logarithmic form gives: $\frac{2x^2 + 9x - 5}{x^2 + 2x - 15} = e^{1} = e.$

3. Cross-multiply:

   • This leads to: $2x^2 + 9x - 5 = e(x^2 + 2x - 15).$

4. Expand and arrange the equation:

   • Thus: $2x^2 + 9x - 5 = ex^2 + 2ex - 15e.$ Rearranging gives: $(2 - e)x^2 + (9 - 2e)x + (15e - 5) = 0.$

5. Use the quadratic formula:

   • From the quadratic formula, we find: $x = \frac{-(9 - 2e) \pm \sqrt{(9 - 2e)^2 - 4(2 - e)(15e - 5)}}{2(2 - e)}.$

Thus, the value of x in terms of e is: $x = \frac{2e - 9 \pm \sqrt{(9 - 2e)^2 - 4(2 - e)(15e - 5)}}{2(2 - e)}.$