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Condense the expression to the logarithm of a single quantity.\(3 \log _{7} x+2 \log _{7} y-4 \log _{7} z\)

Short Answer

Expert verified
The expression simplifies to: \(\log _{7}((x^3 * y^2) / z^{-4})\).

Step by step solution

01

Apply the Power Rule

The power rule of logarithms states that \(a \log_b c = \log_b (c^a)\). Apply this rule to all terms: \(3 \log _{7} x => \log _{7} x^3\), \(2 \log _{7} y => \log _{7} y^2\), and \(-4 \log _{7} z => \log _{7} z^{-4}\). So, the expression now looks as follows: \(\log _{7} x^3 + \log _{7} y^2 - \log _{7} z^{-4}\).
02

Apply the Product Rule

The product rule of logarithms states that \(\log_b a + \log_b c = \log_b (ac)\). Now apply this to the first two terms: \(\log _{7} x^3 + \log _{7} y^2 \) which becomes \(\log _{7} (x^3 * y^2)\). Now the expression simplified to \(\log _{7} (x^3 * y^2) - \log _{7} z^{-4}\).
03

Apply the Quotient Rule

The quotient rule of logarithms states that \(\log_b a - \log_b c = \log_b (a/c)\). Apply this rule to the remaining terms: \(\log _{7} (x^3 * y^2) - \log _{7} z^{-4}\) which simplifies to \(\log _{7}((x^3 * y^2) / z^{-4})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule of Logarithms
Understanding the power rule can significantly simplify solving logarithmic expressions. It translates the multiplication of a logarithm by a number into the power of the logarithm's argument. This is expressed mathematically as: \(a \log_b x = \log_b(x^a)\). To better grasp it, imagine that you have the logarithm of a number, say 2, to the base 10, and it's multiplied by 3. Applying the power rule, this multiplication can be transformed into\( \log_{10}(2^3)\), which is essentially the logarithm of 8, since 2 raised to the power of 3 is 8.

Take for example our exercise: we have terms like \(3 \log_7 x\). When we apply the power rule, it becomes \(\log_7(x^3)\), thus conveying the same information in a more condensed form. This rule is a game-changer in simplifying complex logarithmic expressions into a more manageable format, making it easier to further operate on them with other logarithmic properties.
Product Rule of Logarithms
Moving to the product rule of logarithms, which is invaluable when dealing with the multiplication of variables within logarithmic functions. In essence, it allows us to combine individual logarithms of the same base that are being added into a single logarithm that captures the product of their arguments. Formally, this is shown as \(\log_b a + \log_b c = \log_b (ac)\).

In the context of our exercise, we had two terms\( \log_7 x^3\) and \(\log _7 y^2\). Using the product rule we can combine them, ending up with \(\log_7 (x^3 \cdot y^2)\). This step melds the multiplication of the numbers inside the logarithms and sets up the expression for further simplification. Without the product rule, we would be stuck with separate logs, making it harder to find the final, condensed expression.
Quotient Rule of Logarithms
Last but not least, the quotient rule of logarithms aides in simplifying expressions when division of variables comes into play. It asserts that the difference of two logarithms with the same base can be transformed into the logarithm of a division, mathematically defined as \(\log_b a - \log_b c = \log_b (a/c)\).

Reflecting on our exercise, after applying the product rule we were left with \(\log_7(x^3 \cdot y^2) - \log _7 z^{-4}\). Here's where the quotient rule enters the scene. It lets us combine these two logarithms into one: \(\log_7\left((x^3 \cdot y^2) / z^{-4}\right)\). This property is particularly useful, as it turns a potentially complex subtraction of logs into a single log that represents a fraction, hence dramatically simplifying the calculation process and making the solution more elegant.

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Most popular questions from this chapter

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(5 \log _{10}(x-2)=11\)

The number \(V\) of varieties of suburban nondomesticated wildlife in a community is approximated by the model \(V=15 \cdot 10^{0.02 x}, \quad 0 \leq x \leq 36\) where \(x\) is the number of months since the development of the community was completed. Use this model to approximate the number of months since the development was completed when \(V=50\).

Classify the model as an exponential growth model or an exponential decay model.\(y=3 e^{0.5 t}\)

Super Bowl Ad Cost The table shows the costs \(C\) (in millions of dollars) of a 30 -second TV ad during the Super Bowl for several years from 1987 to \(2006 .\) (Source: TNS Media Intelligence)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Cost } \\ \hline 1987 & 0.6 \\ \hline 1992 & 0.9 \\ \hline 1997 & 1.2 \\ \hline 2002 & 2.2 \\ \hline 2006 & 2.5 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=7\) corresponding to \(1987 .\) (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use a graphing utility to graph the exponential model in base \(e\). (d) Use the exponential model in base \(e\) to predict the costs of a 30 -second ad during the Super Bowl in 2009 and in 2010 .

Compute \(\left[\mathrm{H}^{+}\right]\) for a solution for which \(\mathrm{pH}=7.3\).

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