7. (i) Find the value of $y$ for which $$1.01^{y - 1} = 500$$ Give your answer to 2 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 4

Starting from the equation: $2 \log(3x + 5) = \log(3x + 8) + 1$ We rewrite the right-hand side: $\log(3x + 8) + 1 = \log(3x + 8) + \log(10) = \log(10(3x + 8))$ This allows us to set the logarithms equal: $\log(3x + 5)^2 = \log(10(3x + 8))$ Removing the logarithms: $(3x + 5)^2 = 10(3x + 8)$ Expanding both sides: $9x^2 + 30x + 25 = 30x + 80$ Now, simplifying gives: $9x^2 + 25 - 80 = 0$ Thus: $9x^2 + 18x - 7 = 0$

From the previous step, we have the quadratic equation: $9x^2 + 18x - 7 = 0$ To solve it, we apply the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 9$, $b = 18$, and $c = -7$. Calculating the discriminant: $D = 18^2 - 4 \cdot 9 \cdot (-7) = 324 + 252 = 576$ Since $D$ is a perfect square, using the quadratic formula gives: $x = \frac{-18 \pm \sqrt{576}}{18} = \frac{-18 \pm 24}{18}$ This results in two possible solutions:

1. $x = \frac{6}{18} = \frac{1}{3}$
2. $x = \frac{-42}{18} = -\frac{7}{3}$ Given the constraint $x > -\frac{5}{3}$, the valid solution is: $x = \frac{1}{3}.$