The problem consists of two parts: (a) Solve the logarithmic equation $(y-1) \log_{10} 4 = y \log_{10} 16$ for $y$. (b) Calculate the distance between the house and office, given the walking speeds and the time difference.

1. Problem Description

The problem consists of two parts:

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

(b) Calculate the distance between the house and office, given the walking speeds and the time difference.

2. Solution Steps

(a) Solving the logarithmic equation:

We are given $(y-1) \log_{10} 4 = y \log_{10} 16$.

Since $16 = 4^2$, we have $\log_{10} 16 = \log_{10} 4^2 = 2 \log_{10} 4$.

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

Dividing both sides by $\log_{10} 4$ (since $\log_{10} 4 \ne 0$), we get $y-1 = 2y$.

Subtracting $y$ from both sides, we have $-1 = y$.

So, $y = -1$.

(b) Calculating the distance:

Let $d$ be the distance between the house and the office (in km).

Let $t_1$ be the time taken to walk at 4 km/h, and $t_2$ be the time taken to walk at 5 km/h (in hours).

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 = 0.5 hours.

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

Multiplying both sides by 20 (the least common multiple of 4, 5, and 2), we get $5d = 4d + 10$.

Subtracting $4d$ from both sides, we get $d = 10$.

So, the distance between the house and the office is 10 km.

3. Final Answer

(a) $y = -1$

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