Full solution

Q. Find the numerical value of the log expression.

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\begin{array}{c} \log a=9 \quad \log b=2 \quad \log c=-6 \\ \log \frac{\sqrt[3]{c^{5}}}{a^{9} b^{4}} \end{array}

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Answer:

Apply Quotient Rule: Let's start by expressing the given logarithm using the properties of logarithms. $\newline$ We have $\log \left(\frac{\sqrt[3]{c^{5}}}{a^{9}b^{4}}\right)$ . $\newline$ First, we apply the quotient rule of logarithms, which states that $\log\left(\frac{a}{b}\right) = \log(a) - \log(b)$ .
Rewrite as Difference: Now we rewrite the expression as a difference of two logarithms: $\log(\sqrt[3]{c^{5}}) - \log(a^{9}b^{4})$ .
Apply Power Rule: Next, we apply the power rule of logarithms, which states that $\log(a^n) = n\log(a)$ , to the terms inside the logarithms. $\newline$ For the first term, since $\sqrt[3]{c^{5}}$ is the same as $c^{\frac{5}{3}}$ , we get $\log(c^{\frac{5}{3}}) = \frac{5}{3}\log(c)$ . $\newline$ For the second term, we have two variables inside the logarithm, so we apply the product rule of logarithms, $\log(ab) = \log(a) + \log(b)$ , before applying the power rule.
Apply Product Rule: Applying the product rule to the second term gives us $\log(a^{9}) + \log(b^{4})$ . Now we apply the power rule to each of these terms to get $9\log(a)$ and $4\log(b)$ .
Distribute Negative Sign: Our expression now looks like this: $\newline$ $(\frac{5}{3})\log(c) - (9\log(a) + 4\log(b))$ . $\newline$ We distribute the negative sign through the parenthesis to get: $\newline$ $(\frac{5}{3})\log(c) - 9\log(a) - 4\log(b)$ .
Substitute Given Values: Now we substitute the given values of $\log a$ , $\log b$ , and $\log c$ into the expression. $\newline$ This gives us $(\frac{5}{3})*(-6) - 9*9 - 4*2$ .
Perform Arithmetic Operations: We perform the arithmetic operations: $\frac{5}{3}\times(-6) = -10$ , $9\times9 = 81$ , and $4\times2 = 8$ . So the expression simplifies to \$$-10$ - $81$ - $8$\$.
Find Numerical Value: Finally, we add the numbers together to find the numerical value of the expression: $-10 - 81 - 8 = -99$.