The problem has two parts: (a) Given the equation $(y - 1) \log_{10}4 = y \log_{10}16$, find the value of $y$. (b) If walking at 4 km/h results in arriving 30 minutes later than walking at 5 km/h, calculate the distance between the house and office.

1. Problem Description

The problem has two parts:

(a) Given the equation $(y - 1) \log_{10}4 = y \log_{10}16$, find the value of $y$.

(b) If walking at 4 km/h results in arriving 30 minutes later than walking at 5 km/h, calculate the distance between the house and office.

2. Solution Steps

(a) Solve for $y$ in the equation $(y - 1) \log_{10}4 = y \log_{10}16$.

We know that $16 = 4^2$, so $\log_{10}16 = \log_{10}4^2 = 2 \log_{10}4$.

Substituting this into the equation gives $(y - 1) \log_{10}4 = y (2 \log_{10}4)$.

$(y - 1) \log_{10}4 = 2y \log_{10}4$.

Since $\log_{10}4 \ne 0$, we can divide both sides by $\log_{10}4$ :

$y - 1 = 2y$

Subtract $y$ from both sides:

$-1 = y$

Thus, $y = -1$.

(b) Calculate the distance between the house and the office.

Let $d$ be the distance between the house and the office in kilometers.

Let $t_1$ be the time taken in hours when walking at 4 km/h.

Let $t_2$ be the time taken in hours when walking at 5 km/h.

We know that time = distance / speed.

So, $t_1 = \frac{d}{4}$ and $t_2 = \frac{d}{5}$.

We are given that $t_1 = t_2 + \frac{30}{60}$, since 30 minutes is 0.5 hours.

So, $\frac{d}{4} = \frac{d}{5} + \frac{1}{2}$.

Multiply both sides by 20 to eliminate the fractions:

$20(\frac{d}{4}) = 20(\frac{d}{5} + \frac{1}{2})$

$5d = 4d + 10$

Subtract $4d$ from both sides:

$d = 10$

Thus, the distance is 10 km.

3. Final Answer

(a) $y = -1$

(b) The distance between the house and office is 10 km.