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Use the properties of logarithms to expand the expression as a sum, difference, and/or multiple of logarithms. (Assume all variables are positive.)\(\ln \sqrt{a-1}, \quad a>1\)

Short Answer

Expert verified
The expanded form of the expression using the properties of logarithms is \(\frac{1}{2} \ln (a-1)\).

Step by step solution

01

Express Square Root as Power of 0.5

Rewrite the square root in the expression as a power of \(\frac{1}{2}\). Therefore, the expression \(\ln \sqrt{a-1}\) can be written as \(\ln (a-1)^{\frac{1}{2}}\).
02

Application of Logarithm Power Rule

The power rule states that \(\log_b (m^n) = n \log_b m\). Hence, \(\ln (a-1)^{\frac{1}{2}}\) can be rewritten as \(\frac{1}{2} \ln (a-1)\).
03

Final Simplified Expression

After applying the properties as described, the final simplified expression is \(\frac{1}{2} \ln (a-1)\)

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Key Concepts

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

Logarithm Expansion
When you first encounter a logarithmic function, it can seem a bit like a secret code. However, there are certain properties and rules, like the expansion of logarithms, that make these expressions more manageable. Logarithm expansion is the process of breaking down a logarithm into simpler parts, using specific properties to express it as a sum, difference, or product of logarithms.
In our example, we start with the expression \(\ln \sqrt{a-1}\). The key here is to notice that the square root can be expressed as a power, specifically \(a-1\) raised to the power of 0.5. Understanding this allows us to rewrite the logarithm in a different, often simplified form.
  • Converting roots to fractional powers.
  • Expressing logarithms based on these power rules.
By converting \(\sqrt{a-1}\) into \((a-1)^{\frac{1}{2}}\), the expression became more accessible for further manipulation.
Power Rule
The Power Rule is a fundamental tool in logarithms that allows us to move exponents out in front of the logarithm. The Power Rule states \(\log_b (m^n) = n \log_b m\).
This is handy when you wish to simplify expressions that involve powers, as it trades a power for a coefficient, which is often easier to deal with in calculations.
In the example \(\ln (a-1)^{\frac{1}{2}}\), the Power Rule is applied by moving the exponent \(\frac{1}{2}\) in front of the logarithm:
\[\ln (a-1)^{\frac{1}{2}} = \frac{1}{2} \ln (a-1)\]
  • Exponent becomes a coefficient.
  • Simplifies manipulation of logarithms.
This transformation makes calculations more straightforward and sets you up for smooth subsequent steps in problem-solving.
Simplification of Logarithmic Expressions
After applying logarithmic identities, simplification is often the ultimate goal. Simplifying a logarithmic expression means rewriting it in its most compact and straightforward form, which often reveals more about the relationships within an expression.
In our specific example, once \(\ln (a-1)^{\frac{1}{2}}\) is transformed using the Power Rule, we end with \(\frac{1}{2} \ln (a-1)\).
  • Final expression is simpler and cleaner.
  • Reveals the basic structure and relations.
This type of simplification is especially useful in solving equations, integrating calculus problems, or when logarithmic expressions need to be combined or compared.

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Most popular questions from this chapter

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln x+\ln (x+3)=1\)

Population The populations \(P\) of the United States (in thousands) from 1990 to 2005 are shown in the table. (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1990 & 250,132 \\ \hline 1991 & 253,493 \\ \hline 1992 & 256,894 \\ \hline 1993 & 260,255 \\ \hline 1994 & 263,436 \\ \hline 1995 & 266,557 \\ \hline 1996 & 269,667 \\ \hline 1997 & 272,912 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1998 & 276,115 \\ \hline 1999 & 279,295 \\ \hline 2000 & 282,403 \\ \hline 2001 & 285,335 \\ \hline 2002 & 288,216 \\ \hline 2003 & 291,089 \\ \hline 2004 & 293,908 \\ \hline 2005 & 296,639 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=0\) corresponding to 1990 . (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 the regression feature of a graphing utility to find a linear model and a quadratic model for the data. (d) Use a graphing utility to graph the exponential model in base \(e\) and the models in part (c) with the scatter plot. (e) Use each model to predict the populations in 2008 , 2009 , and 2010 . Do all models give reasonable predictions? Explain.

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\ln x+\ln (x-2)=1\)

Intel The sales per share \(S\) (in dollars) for Intel from 1992 to 2005 can be approximated by the function \(S=\left\\{\begin{array}{lr}-1.48+2.65 \ln t, & 2 \leq t \leq 10 \\ 0.1586 t^{2}-3.465 t+22.87, & 11 \leq t \leq 15\end{array}\right.\) where \(t\) represents the year, with \(t=2\) corresponding to 1992\. (Source: Intel)

A grape has a pH of \(3.5\), and baking soda has a pH of \(8.0\). The hydrogen ion concentration of the grape is how many times that of the baking soda?

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