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

1. Start by rewriting the equation:

   $2\log(3x + 5) = \log(3x + 8) + \log(10)$

2. Use the properties of logarithms to combine the logs on the right:

   $2\log(3x + 5) = \log(10(3x + 8))$

3. Convert the left side using the power rule:

   $\log((3x + 5)^2) = \log(10(3x + 8))$

4. Since the logarithms are equal, set the arguments equal:

   $(3x + 5)^2 = 10(3x + 8)$

5. Expand both sides:

   $9x^2 + 30x + 25 = 30x + 80$

6. Simplify the equation:

   $9x^2 + 25 - 80 = 0$

   $9x^2 + 18x - 7 = 0$

Thus, the equation has been shown.

1. From the previous part, we have the quadratic equation:

   $9x^2 + 18x - 7 = 0$

2. Use the quadratic formula to solve for $x$:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   where $a = 9$, $b = 18$, and $c = -7$:

   $x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 9 \cdot (-7)}}{2 \cdot 9}$

3. Calculate the discriminant and simplify:

   $x = \frac{-18 \pm \sqrt{324 + 252}}{18} = \frac{-18 \pm \sqrt{576}}{18}$

   $x = \frac{-18 \pm 24}{18}$

4. Calculate the two possible values for $x$:

   1st solution: $x = \frac{6}{18} = \frac{1}{3}$

   2nd solution: $x = \frac{-42}{18} = -\frac{7}{3}$

5. Check which values are valid based on $x > -\frac{5}{3}$:

The valid solution is: $x = \frac{1}{3}$