/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 Simplify. $$10^{\log w}$$... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 71

Simplify. $$10^{\log w}$$

Short Answer

Expert verified
The answer is \( w \).

Step by step solution

01

Understanding the problem

We need to simplify the expression \(10^{\log w}\). This involves understanding the properties of logarithms and exponents.
02

Recognizing the property of logarithms

Recall the property that states if \( b eq 1 \) and \( b > 0 \), then \( b^{\log_b x} = x \). In our problem, \( b = 10 \), so \( 10^{\log_{10} w} = w \).
03

Applying the property

Using the property of logarithms, we recognize that \( 10^{\log_{10} w} \) simplifies directly to \( w \). This is because the base of the exponent matches the base of the logarithm.

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Key Concepts

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

Logarithms
Logarithms are a mathematical function that help us deal with exponential equations. They are essentially the opposite of exponents. If you have an equation like 10 raised to some power, the logarithm helps you find that power. For instance, in the expression \(10^{\text{log}_{10} w}\), the \(\text{log}_{10}\) represents the logarithm base 10. This means we are looking for the number that gives us \(w\) when used as an exponent with base 10. Understanding this property is crucial for simplifying equations involving logarithms and exponents.
Exponents
Exponents represent repeated multiplication. For example, \(10^2\) means multiplying 10 by itself, which equals 100. In the context of our problem, we have \(10^{\text{log}_{10} w}\). When the base of an exponent (in this case 10) matches the base of the logarithm (also 10), a special property applies. The property says: if \(b eq 1\) and \(b > 0\), then \(b^{\text{log}_b x} = x\). So, \(10^{\text{log}_{10} w}\) simplifies directly to \(w\). This means the exponents and logarithms cancel each other out, leaving us with a much simpler form.
Simplification
Simplification in mathematics means making an expression easier to work with. In our problem, we are simplifying \(10^{\text{log}_{10} w}\). By using the properties of exponents and logarithms, we recognize that the equation simplifies directly to \(w\). Here's why: the base of the exponent (10) is the same as the base of the logarithm (10). According to the rule, \(b^{\text{log}_b x} = x\), this simplification makes our job easier. No more complicated expressions, just a neat and straightforward result: \(w\). Always look out for opportunities to use these properties to simplify your work.

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Most popular questions from this chapter

Retail e-commerce holiday season sales (November and December), in billions of dollars, \(x\) years after 2009 is given by the function $$ H(x)=4.798 x+28.1778 $$ (Source: statista.com). a) Determine the total amount of e-commerce holiday sales in 2012 and in 2016 . b) Find \(H^{-1}(x)\) and explain what it represents. (PICTURE CANNOT COPY)

Compound Interest. Suppose that \(\$ 82.000\) is invested at \(4 \frac{1}{2} \%\) interest, compounded quarterly. a) Find the function for the amount to which the investment grows after \(t\) years. b) Graph the function. c) Find the amount of money in the account at \(t=0,2\) \(5,\) and 10 years. d) When will the amount of money in the account reach \(\$ 100,000 ?\)

Sketch the graph of the function and check the graph with a graphing calculator. Describe how each graph can be obtained from the graph of a basic exponential function. $$f(x)=1-e^{-0.01 x}$$

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{4}(x+3)+\log _{4}(x-3)=2$$

Determine whether each of the following is true or false. Assume that \(a, x, M,\) and \(N\) are positive. $$\frac{\log _{a} M}{\log _{a} N}=\log _{a} M-\log _{a} N$$
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