Problem 22
Question
In Exercises \(21-30,\) write each logarithm as a sum and\or difference of logarithmic expressions. Eliminate exponents and radicals and evaluate logarithms wherever possible. Assume that \(a, x, y\) \(z>0\) and \(a \neq 1\). $$\log \frac{x^{5} y^{4}}{1000}$$
Step-by-Step Solution
Verified Answer
The logarithmic expression \(\log \frac{x^{5} y^{4}}{1000}\) can be written as a sum and/or difference of logarithmic expressions as \(5 \log \left( {x} \right) + 4 \log \left( {y} \right) - 3\)
1Step 1: Break Down Using Logarithm Properties
Use the quotient property of logarithms to break down the given logarithm into a difference of two terms: \[ \log \left(\frac{x^{5} y^{4}}{1000}\right) = \log \left( {x^{5} y^{4}} \right) - \log \left( {1000} \right) \]
2Step 2: Break Down Further Using Logarithm Properties
Use the product property of logarithms to break down the first term further into a sum of logs which results in: \[ \log \left( {x^{5} y^{4}} \right) - \log \left( {1000} \right) = \log \left( {x^{5}} \right) + \log \left( {y^{4}} \right) - \log \left( {1000} \right) \]
3Step 3: Use the Power Property of Logarithms
Now you can use the power property of logarithms to transfer the power of each log term down as a coefficient and simplify: \[ \log \left( {x^{5}} \right) + \log \left( {y^{4}} \right) - \log \left( {1000} \right) = 5 \log \left( {x} \right) + 4 \log \left( {y} \right) - 3 \]
Key Concepts
Quotient Property of LogarithmsProduct Property of LogarithmsPower Property of Logarithms
Quotient Property of Logarithms
The quotient property of logarithms is a tool you can use when you're dealing with the log of a division. To put it simply, if you have \( \log\left(\frac{a}{b}\right) \) and you want to break it down, the quotient property lets you split it into two separate log terms by subtracting. It becomes \( \log(a) - \log(b) \).
For instance, in our exercise, the original expression was \( \log \left(\frac{x^{5} y^{4}}{1000}\right) \). By applying the quotient property, you divide the problem into a more manageable form: \( \log(x^{5} y^{4}) - \log(1000) \). This property is incredibly useful because it simplifies complex log expressions that have division into smaller, more solvable pieces, making the whole process of understanding and solving logarithms much more approachable.
When using this property, remember that both 'a' and 'b' must be greater than zero because logs are not defined for non-positive values.
For instance, in our exercise, the original expression was \( \log \left(\frac{x^{5} y^{4}}{1000}\right) \). By applying the quotient property, you divide the problem into a more manageable form: \( \log(x^{5} y^{4}) - \log(1000) \). This property is incredibly useful because it simplifies complex log expressions that have division into smaller, more solvable pieces, making the whole process of understanding and solving logarithms much more approachable.
When using this property, remember that both 'a' and 'b' must be greater than zero because logs are not defined for non-positive values.
Product Property of Logarithms
Moving on to the product property of logarithms, this one comes into play when you are dealing with a multiplication within a logarithm. Simplified, if there's an expression like \( \log(ab) \), where 'a' and 'b' are multiplied together under the log, you can separate them into two logs added together - \( \log(a) + \log(b) \).
In our textbook exercise, there was a multiplication inside the log, \( x^{5} y^{4} \). Using the product property, this was separated into \( \log(x^{5}) + \log(y^{4}) \). This makes the expression more straightforward because you can deal with each log individually. It's a way to deconstruct complex logs into something much more digestible, allowing for easier calculation and deeper understanding of the underlying concepts.
Always ensure the values inside the log are positive, as this property doesn't apply to zero or negative values.
In our textbook exercise, there was a multiplication inside the log, \( x^{5} y^{4} \). Using the product property, this was separated into \( \log(x^{5}) + \log(y^{4}) \). This makes the expression more straightforward because you can deal with each log individually. It's a way to deconstruct complex logs into something much more digestible, allowing for easier calculation and deeper understanding of the underlying concepts.
Always ensure the values inside the log are positive, as this property doesn't apply to zero or negative values.
Power Property of Logarithms
Finally, let's talk about the power property of logarithms, another incredibly helpful rule when dealing with powers inside a log. For an expression like \( \log(a^n) \), where 'n' is the exponent, the power property allows you to move the exponent to the front: \( n \cdot \log(a) \).
In the solution to the textbook problem, we had \( \log(x^{5}) \), which according to the power property, became \( 5 \cdot \log(x) \). Similarly, \( \log(y^{4}) \) transformed into \( 4 \cdot \log(y) \). This property is quite powerful because it effectively reduces exponents, which often intimidate students. By utilizing this property, you can turn an exponentiated log into a multiplication problem, leading to significant simplification.
Keep in mind that the original value 'a' in \( \log(a^n) \) must be positive, as logarithms are defined only for positive values, excluding zero.
In the solution to the textbook problem, we had \( \log(x^{5}) \), which according to the power property, became \( 5 \cdot \log(x) \). Similarly, \( \log(y^{4}) \) transformed into \( 4 \cdot \log(y) \). This property is quite powerful because it effectively reduces exponents, which often intimidate students. By utilizing this property, you can turn an exponentiated log into a multiplication problem, leading to significant simplification.
Keep in mind that the original value 'a' in \( \log(a^n) \) must be positive, as logarithms are defined only for positive values, excluding zero.
Other exercises in this chapter
Problem 22
Use \(f(x)=3 \ln x-4\). For what value of \(x\) will \(f(x)=3 ?\)
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State whether each function given by a table is one-to-one. Explain your reasoning. $$\begin{aligned}&x \quad f(x)\\\&\begin{array}{cc}-2 & -9 \\\\-1 & -8\\\0&-
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Evaluate each expression without using a calculator. $$\ln \frac{1}{e}$$
View solution Problem 22
Sketch the graph of each function. $$f(x)=4(2)^{-x}$$
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