Problem 85
Question
In Exercises 81–100, evaluate or simplify each expression without using a calculator. $$ 10^{\log 33} $$
Step-by-Step Solution
Verified Answer
The simplified form of the expression \(10^{\log 33}\) is 33.
1Step 1: Understand the Logarithmic Property
There is a property of logarithms that if \(a = b^{\log_{b} a}\), then \(a\) equals the argument of the logarithm when the base used for the exponentiation and the base of the logarithm are the same. Here, both are base 10.
2Step 2: Apply the Property
For the given expression \(10^{\log 33}\), apply the previously mentioned property. This simplifies the expression to the argument of the logarithm, which is 33 in this case.
Key Concepts
Logarithmic PropertiesSimplifying ExpressionsExponents and Logarithms
Logarithmic Properties
Logarithmic functions have unique rules that make them incredibly useful in math and science. These rules are known as logarithmic properties.
One fundamental property deals with the relationship between exponents and logarithms when they share the same base.
Specifically, if an expression is written in the form \(a^{\log_{a} b}\), then it simplifies to the value of \(b\). This is because taking the logarithm of a number essentially "undoes" the exponentiation when the bases are the same.
In the case of the expression \(10^{\log 33}\), we recognize that it perfectly fits this property, where both the exponent's base and the logarithm's base are 10. This allows us to simplify the expression directly to 33.
One fundamental property deals with the relationship between exponents and logarithms when they share the same base.
Specifically, if an expression is written in the form \(a^{\log_{a} b}\), then it simplifies to the value of \(b\). This is because taking the logarithm of a number essentially "undoes" the exponentiation when the bases are the same.
In the case of the expression \(10^{\log 33}\), we recognize that it perfectly fits this property, where both the exponent's base and the logarithm's base are 10. This allows us to simplify the expression directly to 33.
- This property ensures that calculations involving complicated exponentiating and logarithmic processes are made more manageable.
- It removes the need for computational tools, offering a straightforward solution.
Simplifying Expressions
Simplifying expressions is an essential skill, particularly when dealing with logarithms.
In mathematical terms, simplification means to reduce an expression to its simplest form, making it easier to work with or understand. In the expression \(10^{\log 33}\), you can see it involves an exponent and a logarithm with matching bases.
To simplify expressions like this:
In mathematical terms, simplification means to reduce an expression to its simplest form, making it easier to work with or understand. In the expression \(10^{\log 33}\), you can see it involves an exponent and a logarithm with matching bases.
To simplify expressions like this:
- Identify if the expression fits any known properties, such as the one where \(b^{\log_{b}a}=a\).
- Apply the property to remove unnecessary components and reach a straightforward result, as we did with 33.
Exponents and Logarithms
Exponents and logarithms are core concepts in algebra that have a reciprocal relationship. Understanding how they interact can make it easier to handle problems involving both.
An exponent shows how many times we use a number in a multiplication, labeled as the base raised to a power. Conversely, a logarithm answers the question: "To what power must we raise a given base to get a specific number?"
For instance, in the equation \(b^{x} = y\), \(x\) would be \(\log_{b} y\).
This interchange allows for converting exponential expressions to logarithmic form and vice versa, streamlining the process of solving complex equations.
An exponent shows how many times we use a number in a multiplication, labeled as the base raised to a power. Conversely, a logarithm answers the question: "To what power must we raise a given base to get a specific number?"
For instance, in the equation \(b^{x} = y\), \(x\) would be \(\log_{b} y\).
This interchange allows for converting exponential expressions to logarithmic form and vice versa, streamlining the process of solving complex equations.
- This means exponentials and logarithms act as inverse operations. If you take the logarithm of a base-raised-to-a-power, you return to the power.
- In a practical sense, recognizing these interactions helps in fields like data analysis, finance, and natural sciences.
Other exercises in this chapter
Problem 85
Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\)Write each expression in terms of \(A\) and \(C\). $$ \log _{b} 8 $$
View solution Problem 85
Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer.
View solution Problem 85
determine whether each statement makes sense or does not make sense, and explain your reasoning. Taxing thoughts: I'm looking at data that show the number of pa
View solution Problem 86
Exercises \(86-88\) will help you prepare for the material covered in the first section of the next chapter. $$ \text { Solve: } \quad \frac{5 \pi}{4}=2 \pi x $
View solution