We are given two problems: (a) We need to find the value of $y$ given the equation $(y-1) \log_{10}4 = y \log_{10}16$. (b) We need to find the distance between the house and the office, given that walking at 4 km/h results in arriving 30 minutes later than walking at 5 km/h.

1. Problem Description

We are given two problems:

(a) We need to find the value of $y$ given the equation $(y-1) \log_{10}4 = y \log_{10}16$.

(b) We need to find the distance between the house and the office, given that walking at 4 km/h results in arriving 30 minutes later than walking at 5 km/h.

2. Solution Steps

(a) Solving the logarithmic equation:

We have the equation $(y-1) \log_{10}4 = y \log_{10}16$.

We can rewrite $\log_{10}16$ as $\log_{10}4^2 = 2\log_{10}4$.

So the equation becomes $(y-1) \log_{10}4 = y (2 \log_{10}4)$.

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

$y-1 = 2y$

$y-2y = 1$

$-y = 1$

$y = -1$

(b) Solving the distance problem:

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 $t_1 = t_2 + 30 \text{ minutes}$. Since we are using hours, we need to convert 30 minutes to hours, which is $\frac{30}{60} = \frac{1}{2}$ hours.

So, $t_1 = t_2 + \frac{1}{2}$.

Substituting the expressions for $t_1$ and $t_2$:

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

To solve for $d$, we can multiply both sides by 20 (the least common multiple of 4, 5, and 2):

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

$5d = 4d + 10$

$5d - 4d = 10$

$d = 10$

3. Final Answer

(a) $y = -1$

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