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Chapter 3: Problem 21

Evaluate the given quantities assuming that $$ \begin{array}{l} \log _{3} x=5.3 \text { and } \log _{3} y=2.1 \\ \log _{4} u=3.2 \text { and } \log _{4} v=1.3 \end{array} $$ $$ \log _{3} \frac{x}{3 y} $$

Short Answer

Expert verified
\(\log_{3} \frac{x}{3y} = 2.2\)

Step by step solution

01

Convert logarithmic equations to exponential equations

Using the identities \(\log_a b = c\) is equivalent to \(a^c = b\), we can convert the given logarithmic equations for x, y, u and v into exponential equations as follows: 1. \(3^{5.3} = x\) 2. \(3^{2.1} = y\) 3. \(4^{3.2} = u\) 4. \(4^{1.3} = v\)
02

Rewrite the required expression using x and y

We are asked to find the value of \(\log_{3} \frac{x}{3y}\). Using the exponential equations from Step 1, we can rewrite this expression as: \(\log_{3} \frac{3^{5.3}}{3(3^{2.1})}\)
03

Simplify the fraction inside the logarithm

Now, we'll simplify the fraction inside the logarithm using the properties of exponents: \(\log_{3} \frac{3^{5.3}}{3(3^{2.1})} = \log_{3} \frac{3^{5.3}}{3^{1}(3^{2.1})}\) Remember that when multiplying numbers with the same base, we add their exponents: \(\log_{3} \frac{3^{5.3}}{3^{1}(3^{2.1})} = \log_{3} \frac{3^{5.3}}{3^{1+2.1}}\) \(\log_{3} \frac{3^{5.3}}{3^{1+2.1}} = \log_{3} \frac{3^{5.3}}{3^{3.1}}\) Now, when dividing numbers with the same base, we subtract their exponents: \(\log_{3} \frac{3^{5.3}}{3^{3.1}} = \log_{3}(3^{5.3 - 3.1}) = \log_{3}(3^{2.2})\)
04

Find the logarithm

Since the base and the argument are the same, the logarithm is simply the exponent: \(\log_{3}(3^{2.2}) = 2.2\) So, \(\log_{3} \frac{x}{3y} = 2.2\).

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