Problem 86
Question
In Exercises 81–100, evaluate or simplify each expression without using a calculator. $$ 10^{\log 53} $$
Step-by-Step Solution
Verified Answer
The simplified expression of \(10^{log 53}\) is 53.
1Step 1 - Identify the base
Identify the base of the exponent and the logarithm. We can see that both the logarithm and the exponent have the same base, which is 10.
2Step 2 - Use the inverse property of logarithms
Using the inverse property of logarithms, \(a^{log_a b}=b\), that says if a number raised to the power of a logarithm with the same base results in the argument of the logarithm. The base 'a' is 10 in this equation, so we can simplify \(10^{log 53}\) to 53. This is essentially calculating 10 raised to the power that gives us 53, which is just 53 itself.
Key Concepts
Inverse Property of LogarithmsExponential FunctionsSimplification of Expressions
Inverse Property of Logarithms
Logarithms and exponents are closely related mathematical concepts. The inverse property of logarithms is a key principle that greatly simplifies certain expressions. It's similar to undoing an operation, just like adding and subtracting are inverse operations. This property states that if you have a base raised to a logarithm of a number with the same base, you simply end up with that number. In mathematical terms, this is written as:
- \( a^{\log_a b} = b \)
- \( 10^{\log 53} = 53 \).
Exponential Functions
Exponential functions are functions where a constant base is raised to a variable exponent. They are of the form:
Exponential functions often model situations of growth or decay, such as population growth or radioactive decay.
Using exponentials with logarithms, we can find the power needed to achieve a certain number, particularly when both use the same base.In the exercise, the base and the logarithm's base match, which informs how we apply the inverse property of logarithms. This interplay highlights why recognizing this relationship is crucial for simplification.
- \( y = a^x \)
Exponential functions often model situations of growth or decay, such as population growth or radioactive decay.
Using exponentials with logarithms, we can find the power needed to achieve a certain number, particularly when both use the same base.In the exercise, the base and the logarithm's base match, which informs how we apply the inverse property of logarithms. This interplay highlights why recognizing this relationship is crucial for simplification.
Simplification of Expressions
When we talk about simplifying expressions involving logarithms and exponentials, our goal is to make the expression as straightforward as possible.
Instead of blindly applying rules, it's important to see these shortcuts as tools to reveal simpler paths to your solution.
In our activity, the expression \(10^{\log 53}\) directly simplified to 53 using its properties, saving time and reducing the potential for errors in manual calculations.
- Look for opportunities to apply properties like the inverse property of logarithms, which can greatly shrink the apparent complexity of a problem.
- Check whether the base of the exponent and the base of the logarithm match. If they do, you might be able to use the inverse property.
Instead of blindly applying rules, it's important to see these shortcuts as tools to reveal simpler paths to your solution.
In our activity, the expression \(10^{\log 53}\) directly simplified to 53 using its properties, saving time and reducing the potential for errors in manual calculations.
Other exercises in this chapter
Problem 86
Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\)Write each expression in terms of \(A\) and \(C\). $$ \log _{b} 81 $$
View solution Problem 86
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 86
determine whether each statement makes sense or does not make sense, and explain your reasoning. I use the natural base \(e\) when determining how much money I'
View solution Problem 87
Exercises \(86-88\) will help you prepare for the material covered in the first section of the next chapter. $$ \text { Simplify: } \frac{17 \pi}{6}-2 \pi $$
View solution