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\(19-44\) Use the Laws of Logarithms to expand the expression. $$ \ln \frac{3 x^{2}}{(x+1)^{10}} $$

Short Answer

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\(\ln 3 + 2\ln x - 10\ln (x+1)\)

Step by step solution

01

Apply the Quotient Rule of Logarithms

The quotient rule of logarithms states that \(\ln \left(\frac{a}{b}\right) = \ln a - \ln b\). Apply this to the expression:\[\ln \frac{3x^2}{(x+1)^{10}} = \ln (3x^2) - \ln ((x+1)^{10})\]
02

Expand the Product Inside the Logarithm

Use the product rule of logarithms, which states that \(\ln (ab) = \ln a + \ln b\), to expand \(\ln (3x^2)\):\[\ln (3x^2) = \ln 3 + \ln x^2\]
03

Apply the Power Rule of Logarithms

Apply the power rule \(\ln (a^b) = b \ln a\) to both \(\ln x^2\) and \(\ln ((x+1)^{10})\):- For \(\ln x^2\):\[\ln x^2 = 2 \ln x\]- For \(\ln ((x+1)^{10})\):\[\ln ((x+1)^{10}) = 10 \ln (x+1)\]
04

Combine All Parts

Substitute all expansions back into the expression:\[\ln \frac{3x^2}{(x+1)^{10}} = (\ln 3 + 2\ln x) - (10 \ln (x+1))\]Finally, combine and simplify:\[\ln 3 + 2\ln x - 10\ln (x+1)\]

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Key Concepts

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

Quotient Rule of Logarithms
The Quotient Rule of Logarithms is a fundamental principle that simplifies the logarithm of a quotient into the difference of two logarithms. Think of it as a way to break down division inside a logarithmic expression. According to the rule, if you have a logarithm of a division of two numbers, such as \( \ln \left(\frac{a}{b}\right) \), you can separate that into \( \ln a - \ln b \). This rule helps make calculations or manipulations involving logarithms easier. In our exercise, using this rule transforms \( \ln \frac{3x^2}{(x+1)^{10}} \) into \( \ln (3x^2) - \ln ((x+1)^{10}) \).By keeping the division as a simple subtraction of logs, complex expressions become straightforward to handle. It is particularly useful for solving equations or simplifying mathematical expressions where logarithms are involved. Remember, the quotient rule is your go-to for dealing with divisions in logarithmic form!
Product Rule of Logarithms
The Product Rule of Logarithms is a handy tool for expanding logarithms involving products. It states that the logarithm of a product is the sum of the logarithms of the factors. So, if you have an expression like \( \ln(ab) \), it expands to \( \ln a + \ln b \).In our context, this rule applied to the product \( 3x^2 \) inside the logarithm gives \( \ln 3 + \ln x^2 \). This step helps decompose logarithms of complex or bulky expressions into simpler parts. By using the product rule, you can focus on managing each factor separately, which is much easier than trying to handle all at once.Always remember to break down every component inside the logarithm over multiplication to fully leverage this rule. This simplification can significantly aid in solving and understanding larger mathematical problems.
Power Rule of Logarithms
The Power Rule of Logarithms is all about making exponents in a logarithmic expression easier to manage by "bringing them down". Formally, it says that the logarithm of a power, like \( \ln(a^b) \), can be rewritten as \( b \ln a \).This rule is extremely useful when you deal with expressions where variables are raised to powers. In our problem, applying the power rule to \( \ln x^2 \) and \( \ln((x+1)^{10}) \) results in \( 2 \ln x \) and \( 10 \ln(x+1) \), respectively.By rewriting the exponents as coefficients, you simplify the expression and make it much more manageable. This operation not only simplifies algebraic manipulation but also provides insights into how the variables are contributing to the final value of the logarithm.The power rule is an essential tool when working with logarithmic expressions, especially when expansions or simplifications are needed. It turns potentially complex symbolic representations into straightforward linear operations.

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Most popular questions from this chapter

Each of these equations can be transformed into an equation of linear or quadratic type by applying the hint. Solve each equation. (a) \((x-1)^{\log (x-1)}=100(x-1)\) (b) \(\log _{2} x+\log _{4} x+\log _{8} x=11\) (c) \(4^{x}-2^{x+1}=3\)

Carbon Dating The age of an ancient artifact can be determined by the amount of radioactive carbon-14 remaining in it. If \(D_{0}\) is the original amount of carbon- 14 and \(D\) is the amount remaining, then the artifact's age \(A\) (in years) is given by $$ A=-8267 \ln \left(\frac{D}{D_{0}}\right) $$ Find the age of an object if the amount \(D\) of carbon \(-14\) that remains in the object is 73\(\%\) of the original amount \(D_{0}\) .

(a) By drawing the graphs of the functions $$ f(x)=1+\ln (1+x) \quad \text { and } \quad g(x)=\sqrt{x} $$ in a suitable viewing rectangle, show that even when a logarithmic function starts out higher than a root function, it is ultimately overtaken by the root function. (b) Find, correct to two decimal places, the solutions of the equation \(\sqrt{x}=1+\ln (1+x)\)

Use a graphing device to find all solutions of the equation, rounded to two decimal places. \(2^{-x}=x-1\)

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