A certain radioactive substance is known to decay with time such that its mass, $M$ g, after $t$ hours is given by $M = 100\left(4^{-kt}\right)$, where $k$ is a positive constant. The original mass of a block of substance is 100 g. After 40 hours, its mass decreased to 90 g.

1. Substitute the given values of $M$ and $t$ into the equation

$90 = 100(4^{-40k})$

2. Isolate the exponential term

$4^{-40k} = \frac{90}{100}$
$4^{-40k} = 0.9$

3. Use logarithms to solve for $k$

$\ln(4^{-40k}) = \ln(0.9)$
$-40k \ln(4) = \ln(0.9)$
$k = \frac{\ln(0.9)}{-40 \ln(4)}$
$k \approx 0.00190004$
$k = 0.00190 \text{ (to 3 s.f.)}$

1. Find the value of $k$ from the information given in part (i)

From part (i), when $t = 40$, $M = 90$:
$90 = 100(4^{-40k})$
$0.9 = 4^{-40k}$
$-40k \ln 4 = \ln 0.9$
$k = -\frac{\ln 0.9}{40 \ln 4} \approx 0.0018999$

2. Substitute $t = 263$ and $k$ to find the remaining mass and percentage

When $t = 263$:
$M = 100(4^{-263 \times 0.0018999})$
$M = 100(0.9)^{\frac{263}{40}}$
$M \approx 50.033$
Since the original mass is $100\text{ g}$, the percentage remaining is:
$\frac{50.033}{100} \times 100\% = 50.0\% \text{ (to 3 s.f.)}$

1. Change the base of the second logarithmic term to base 5.

$\log_{5} x + \frac{\log_{5} (x-2)}{\log_{5} 25} = 1$
$\log_{5} x + \frac{\log_{5} (x-2)}{2} = 1$

2. Multiply by 2 and use the power law of logarithms.

$2\log_{5} x + \log_{5} (x-2) = 2$
$\log_{5} x^2 + \log_{5} (x-2) = 2$

3. Combine the logarithms using the product law and convert to exponential form to obtain the cubic equation.

$\log_{5} [x^2(x-2)] = 2$
$x^2(x-2) = 5^2$
$x^3 - 2x^2 = 25$
$x^3 - 2x^2 - 25 = 0$