We are given two problems. (a) We are asked to find the value of $y$ in the equation $(y-1) \log_{10}4 = y \log_{10}16$ without using mathematical tables or a calculator. (b) We are asked to calculate the distance between a house and an 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 are asked to find the value of $y$ in the equation $(y-1) \log_{10}4 = y \log_{10}16$ without using mathematical tables or a calculator.

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

2. Solution Steps

(a)

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

We can write $4$ as $2^2$ and $16$ as $2^4$. Thus, we have:

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

Using the logarithm property $\log a^b = b \log a$, we have:

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

Dividing both sides by $2 \log_{10}2$ (since $\log_{10}2 \ne 0$), we get:

$y-1 = 2y$

Subtracting $y$ from both sides, we get:

$-1 = y$

So, $y = -1$.

(b)

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 distance = speed × time. So, time = distance / speed.

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

We are given that walking at 4 km/h takes 30 minutes (or 0.5 hours) longer than walking at 5 km/h.

So, $t_1 = t_2 + 0.5$.

Substituting the expressions for $t_1$ and $t_2$, we have:

$\frac{d}{4} = \frac{d}{5} + 0.5$

Multiplying both sides by 20 (the least common multiple of 4 and 5), 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.