The problem contains three parts: (a) Evaluate $(0.004592)^{\frac{1}{3}}$ using logarithm tables. (b) Express $y$ in terms of $x$ given that $\log_{10} y + 3\log_{10} x = 2$. (c) Solve the system of linear equations: $3x - 2y = 21$ $4x + 5y = 5$

1. Problem Description

The problem contains three parts:

(a) Evaluate $(0.004592)^{\frac{1}{3}}$ using logarithm tables.

(b) Express $y$ in terms of $x$ given that $\log_{10} y + 3\log_{10} x = 2$.

(c) Solve the system of linear equations:

$3x - 2y = 21$

$4x + 5y = 5$

2. Solution Steps

(a) Let $N = (0.004592)^{\frac{1}{3}}$. Then, $\log_{10} N = \frac{1}{3} \log_{10} 0.004592$.

First, find $\log_{10} 0.004592$. We can write $0.004592 = 4.592 \times 10^{-3}$.

Then $\log_{10} 0.004592 = \log_{10} (4.592 \times 10^{-3}) = \log_{10} 4.592 + \log_{10} 10^{-3} = \log_{10} 4.592 - 3$.

From logarithm tables, $\log_{10} 4.592 \approx 0.6620$.

Therefore, $\log_{10} 0.004592 \approx 0.6620 - 3 = -2.338$. We can write this as $\bar{3}.6620$.

$\log_{10} N = \frac{1}{3} \log_{10} 0.004592 \approx \frac{1}{3} (\bar{3}.6620) = \frac{1}{3} (-3 + 0.6620) = -1 + 0.2207 = \bar{1}.2207$.

$N = \text{antilog}_{10} (\bar{1}.2207)$.

From antilogarithm tables, the antilog of $0.2207$ is approximately $1.662$.

So, $N = 1.662 \times 10^{-1} = 0.1662$.

(b) $\log_{10} y + 3\log_{10} x = 2$

$\log_{10} y + \log_{10} x^3 = 2$

$\log_{10} (yx^3) = 2$

$yx^3 = 10^2$

$yx^3 = 100$

$y = \frac{100}{x^3}$

(c)

$3x - 2y = 21$ (1)

$4x + 5y = 5$ (2)

Multiply (1) by 5 and (2) by 2:

$15x - 10y = 105$ (3)

$8x + 10y = 10$ (4)

Add (3) and (4):

$23x = 115$

$x = \frac{115}{23} = 5$

Substitute $x=5$ into (1):

$3(5) - 2y = 21$

$15 - 2y = 21$

$-2y = 6$

$y = -3$

Therefore, $x=5$ and $y=-3$.

3. Final Answer

(a) $0.1662$

(b) $y = \frac{100}{x^3}$

(c) $x = 5$, $y = -3$