/*! 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 7 Give the value of each expressio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 7

Give the value of each expression. \(10^{\log \sqrt{3}}\)

Short Answer

Expert verified
\10^{\frac{1}{2} \log \sqrt{3} \)=3^{\frac{1}{2}}\) \sqrt{3}.

Step by step solution

01

Understand the logarithmic property

Recognize that the expression \(10^{\frac{1}{2} \log 3}\) utilizes the property of logarithms that allows us to simplify powers within logarithmic functions.
02

Use the power rule for logarithms

Apply the power rule of logarithms, which states \(a^\log_b(c) = c^{\log_b(a)}\). In this case, express 10 as 10 and place \sqrt{3}\ in the logarithm to match the form: \(10^{\frac{1}{2} \log 10^3} = (10^3)^{\frac{1}{2}}\ or 10^{\frac{1}{2} \log 10^3} = 10^{3/2}= 3^{\frac{1}{2}\) \sqrt{3}.}

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Logarithms
Logarithms are the inverse operations of exponentiation. If you have an exponential equation, such as \(10^x = y\), you can use logarithms to rewrite it as \(x = \log_{10}(y)\).
They help us solve equations where the exponent is unknown.

Some important properties of logarithms include:
  • Logarithm of a product: \(\log_b(xy) = \log_b(x) + \log_b(y)\)
  • Logarithm of a quotient: \(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)
  • Logarithm of a power: \(\log_b(x^y) = y \log_b(x)\)
These properties can simplify complex expressions and make calculations with logs more manageable.
Power Rule of Logarithms
The power rule of logarithms is a tool that many students find handy. It states that \(\log_b(x^y) = y \log_b(x)\).
In simpler terms, you can 'bring down' the exponent as a multiplier in front of the logarithm. This makes it easier to work with logarithmic expressions.

For example, consider the expression in our exercise, \(10^{\frac{1}{2} \log 3}\). Using the power rule:
  • First, note that \(\frac{1}{2}\log 3\) means half the logarithm of 3.
  • We can rewrite it using the power rule: \((\log 3^{1/2})\) because \(1/2\) becomes the exponent of the logarithm.
  • And we know that \(\sqrt{3}\) is another way to write \(3^{1/2}\).
So, our expression eventually simplifies as showing, the essence of the 'power rule of logarithms'.
Simplifying Expressions
Simplifying expressions is a crucial skill in mathematics. The goal is to make an expression as simple as possible, using various algebraic rules.

Let's see an example with the logarithmic expression we started with: \(10^{\frac{1}{2} \log 3}\). Here's the step-by-step:
  • Start by recognizing \(10^{\frac{1}{2} \log 10^3}\). Use the log property \(\log_a(a) = 1\).
  • Apply the power rule: \((10^3)^{1/2}\).
  • Simplify \(10^{3/2}\) which means \(10^{1.5}\).
These steps allow you to break down complex expressions into simpler components, making the math much easier to handle.
Practicing these steps will help you become more comfortable with logarithms and other algebraic principles. And remember, patience and practice are keys to mastering these concepts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve each equation. \(\log _{4} \sqrt{64}=x\)

Solve each equation. Use natural logarithms. Approximate solutions to three decimal places when appropriate. $$ \ln e^{2 x}=\pi $$

Use the special properties of logarithms to evaluate each expression. \(\log _{5} 5^{6}\)

Solve each equation. \(\log _{x} 125=-3\)

To four decimal places, the values of \(\log _{10} 2\) and \(\log _{10} 9\) are $$\log _{10} 2=0.3010 \text { and } \log _{10} 9=0.9542$$ Use these values and the properties of logarithms to evaluate each expression. DO NOT USE A CALCULATOR. $$\log _{10} 36$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.