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Use the properties of logarithms to simplify the expression. \(9^{\log _{9} 15}\)

Short Answer

Expert verified
The simplified form of the given expression \(9^{\log _{9} 15}\) is 15.

Step by step solution

01

Identification of Logarithmic Property

Firstly, identify and comprehend the logarithmic property which states that if \(a = b^c\), then log base \(b\) of \(a\) equals \(c\). This also implies that \(a = b^{\log _{b} a}\). It also means that, if the base of the exponent and the base of the logarithm are the same, they cancel each other out, leaving only the argument of the logarithm behind.
02

Application of Identified Logarithmic Property

Apply the identified logarithmic property to the given expression \(9^{\log _{9} 15}\). The bases of the exponent and the logarithm are the same (which is 9), so according to the identified property, they cancel each other out. The only thing left is the argument of the logarithm, which is 15.

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Key Concepts

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

Logarithmic Identities
Understanding logarithmic identities is crucial to mastering the art of simplifying logarithmic expressions. A logarithm is an exponent to which a specified base must be raised to obtain some number. For example, when we write \(\log_b a = c\), it means that \(b^c = a\). This becomes especially handy with one of the core logarithmic identities: \(b^{\log_b a} = a\), which reveals that the base \(b\) raised to the power of its own log with respect to \(a\) simply results in \(a\).

This identity is a reflection of the inverse relationship between exponents and logarithms, effectively allowing you to switch between the exponential and logarithmic form. Knowing this identity makes it possible to assess expressions like \(9^{\log_9 15}\) effortlessly, seeing that the result will be the argument of the logarithm, which is 15, as the base and the base of the logarithm are identical and, therefore, cancel each other out.
Simplifying Logarithmic Expressions
When faced with the task of simplifying logarithmic expressions, a good understanding of logarithmic properties can turn a seemingly complex problem into a solvable one. Simplification typically involves applying logarithmic identities and rules to break down the expression into a more manageable form.

Common techniques involve using the power rule \(\log_b (a^c) = c\cdot\log_b a\), the product rule \(\log_b (ab) = \log_b a + \log_b b\), and the quotient rule \(\log_b (a/b) = \log_b a - \log_b b\). Additionally, recognizing when a term can be rewritten using these identities is key. For instance, if you encounter \(\log_b b\), it simplifies to 1 because any number raised to the first power is itself. Being able to manipulate these expressions confidently paves the way to solving more intricate logarithmic equations.
Logarithm Base Change
The base change rule for logarithms is a powerful tool when dealing with logarithms of different bases. It states that for any positive numbers \(a\), \(b\), and \(c\), where \(b eq 1\) and \(c eq 1\), you can convert from one base to another using the formula \(\log_b a = \frac{\log_c a}{\log_c b}\). This can be immensely useful when a calculation involves a base that's not convenient for a given situation, like lacking a particular button on a calculator or requiring a common base to combine terms.

Using base change can simplify expressions by harmonizing bases. It's also essential when solving logarithms that require bases to coincide with standard mathematical constants or when comparing multiple logarithmic terms with differing bases. The base change formula is a testimony to the versatility and coherent structure of logarithms across various mathematical contexts.

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Most popular questions from this chapter

Condensing a Logarithmic Expression In Exercises \(67-82,\) condense the expression to the logarithm of a single quantity. $$\log x-2 \log (x+1)$$

Comparing Models A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C} .\) The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$ \begin{array}{l}{\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)} \\\ {\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right)}\end{array} $$ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\) . (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This rasulting line has the form \(\ln (T-21)=a t+b\) . Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$ \left(t, \frac{1}{T-21}\right) $$ Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$ \frac{1}{T-21}=a t+b $$ Solve for \(T,\) and use the graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

Using the Change-of-Base Formula In Exercises \(11-14,\) evaluate the logarithm using the change-of-base formula. Round your result to three decimal places. $$\log _{9} 0.1$$

Using Properties of Logarithms In Exercises \(21-36,\) find the exact value of the logarithmic expression without using a calculator. (If this is not possible, then state the reason.) $$\ln \frac{1}{\sqrt{e}}$$

Using Properties of Logarithms In Exercises \(15-20\) , use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\ln \left(5 e^{6}\right)$$

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