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 9 - 2018 - Paper 4

Starting with the equation:

$2 \log(3x + 5) = \log(3x + 8) + 1$

Using the property of logarithms: $2 \log(3x + 5) = \log((3x + 5)^2)$

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

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

Removing the logarithms gives: $(3x + 5)^2 = 10(3x + 8)$

Expanding both sides: $(3x + 5)^2 = 9x^2 + 30x + 25$ $10(3x + 8) = 30x + 80$

Setting these equal to each other: $9x^2 + 30x + 25 = 30x + 80$

Cancelling $30x$ from both sides leads us to: $9x^2 + 25 - 80 = 0$

This simplifies to: $9x^2 - 55 = 0$

So we can rewrite this as: $9x^2 + 18x - 7 = 0$.

Now that we have established the quadratic equation:

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

we can solve for $x$ using the quadratic formula:

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

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

Calculating the discriminant: $b^2 - 4ac = 18^2 - 4(9)(-7) = 324 + 252 = 576.$

Continuing with the quadratic formula:

$x = \frac{-18 \pm \sqrt{576}}{2 \times 9} = \frac{-18 \pm 24}{18}$

This leads to two potential solutions:

1. $x = \frac{6}{18} = \frac{1}{3}$
2. $x = \frac{-42}{18} = -\frac{7}{3}$

Given the constraint that $x > -\frac{5}{3}$, we conclude: The solution to the original equation is $x = \frac{1}{3}$.