We are given the values of $\log a$, $\log b$, and $\log c$, and we are asked to evaluate four logarithmic expressions involving $a$, $b$, and $c$. Specifically, $\log a = 2$, $\log b = -1$, and $\log c = 1\frac{1}{2} = \frac{3}{2}$. We need to evaluate the following: (a) $\log(a^2bc)$ (b) $\log(\frac{a^2}{bc^3})$ (c) $\log(a \sqrt{\frac{b}{c^3}})$ (d) $\log(\frac{\sqrt{10ac^3}}{b^2})$

1. Problem Description

We are given the values of $\log a$, $\log b$, and $\log c$, and we are asked to evaluate four logarithmic expressions involving $a$, $b$, and $c$.

Specifically, $\log a = 2$, $\log b = -1$, and $\log c = 1\frac{1}{2} = \frac{3}{2}$. We need to evaluate the following:

(a) $\log(a^2bc)$

(b) $\log(\frac{a^2}{bc^3})$

(c) $\log(a \sqrt{\frac{b}{c^3}})$

(d) $\log(\frac{\sqrt{10ac^3}}{b^2})$

2. Solution Steps

(a) $\log(a^2bc)$

Using the logarithm product rule: $\log(xy) = \log x + \log y$

$\log(a^2bc) = \log(a^2) + \log(b) + \log(c)$

Using the power rule: $\log(x^n) = n\log x$

$\log(a^2) = 2\log a$

So, $\log(a^2bc) = 2\log a + \log b + \log c$

Substituting the given values:

$\log(a^2bc) = 2(2) + (-1) + \frac{3}{2} = 4 - 1 + \frac{3}{2} = 3 + \frac{3}{2} = \frac{6}{2} + \frac{3}{2} = \frac{9}{2} = 4.5$

(b) $\log(\frac{a^2}{bc^3})$

Using the logarithm quotient rule: $\log(\frac{x}{y}) = \log x - \log y$

$\log(\frac{a^2}{bc^3}) = \log(a^2) - \log(bc^3)$

Using the logarithm product rule: $\log(bc^3) = \log b + \log(c^3)$

Using the power rule: $\log(c^3) = 3\log c$

$\log(\frac{a^2}{bc^3}) = \log(a^2) - (\log b + 3\log c)$

Using the power rule: $\log(a^2) = 2\log a$

$\log(\frac{a^2}{bc^3}) = 2\log a - \log b - 3\log c$

Substituting the given values:

$\log(\frac{a^2}{bc^3}) = 2(2) - (-1) - 3(\frac{3}{2}) = 4 + 1 - \frac{9}{2} = 5 - \frac{9}{2} = \frac{10}{2} - \frac{9}{2} = \frac{1}{2} = 0.5$

(c) $\log(a \sqrt{\frac{b}{c^3}})$

$\log(a \sqrt{\frac{b}{c^3}}) = \log a + \log(\sqrt{\frac{b}{c^3}})$

Using the power rule: $\log(\sqrt{\frac{b}{c^3}}) = \log((\frac{b}{c^3})^{\frac{1}{2}}) = \frac{1}{2} \log(\frac{b}{c^3})$

Using the logarithm quotient rule: $\log(\frac{b}{c^3}) = \log b - \log(c^3)$

Using the power rule: $\log(c^3) = 3\log c$

$\log(a \sqrt{\frac{b}{c^3}}) = \log a + \frac{1}{2}(\log b - 3\log c)$

Substituting the given values:

$\log(a \sqrt{\frac{b}{c^3}}) = 2 + \frac{1}{2}(-1 - 3(\frac{3}{2})) = 2 + \frac{1}{2}(-1 - \frac{9}{2}) = 2 + \frac{1}{2}(-\frac{2}{2} - \frac{9}{2}) = 2 + \frac{1}{2}(-\frac{11}{2}) = 2 - \frac{11}{4} = \frac{8}{4} - \frac{11}{4} = -\frac{3}{4} = -0.75$

(d) $\log(\frac{\sqrt{10ac^3}}{b^2})$

$\log(\frac{\sqrt{10ac^3}}{b^2}) = \log(\sqrt{10ac^3}) - \log(b^2)$

$\log(\sqrt{10ac^3}) = \log((10ac^3)^{\frac{1}{2}}) = \frac{1}{2} \log(10ac^3)$

$\log(10ac^3) = \log 10 + \log a + \log(c^3)$

Assuming base 10 logarithm, $\log 10 = 1$

$\log(c^3) = 3\log c$

$\log(\frac{\sqrt{10ac^3}}{b^2}) = \frac{1}{2}(1 + \log a + 3\log c) - 2\log b$

Substituting the given values:

$\log(\frac{\sqrt{10ac^3}}{b^2}) = \frac{1}{2}(1 + 2 + 3(\frac{3}{2})) - 2(-1) = \frac{1}{2}(3 + \frac{9}{2}) + 2 = \frac{1}{2}(\frac{6}{2} + \frac{9}{2}) + 2 = \frac{1}{2}(\frac{15}{2}) + 2 = \frac{15}{4} + 2 = \frac{15}{4} + \frac{8}{4} = \frac{23}{4} = 5.75$

3. Final Answer

(a) $\log(a^2bc) = 4.5$

(b) $\log(\frac{a^2}{bc^3}) = 0.5$

(c) $\log(a \sqrt{\frac{b}{c^3}}) = -0.75$

(d) $\log(\frac{\sqrt{10ac^3}}{b^2}) = 5.75$