/*! 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 68 Simplify. $$\log _{q} q^{\sqrt... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 68

Simplify. $$\log _{q} q^{\sqrt{3}}$$

Short Answer

Expert verified
The simplified form is \( \sqrt{3} \ \).

Step by step solution

01

Understand the Logarithmic Property

Familiarize yourself with the logarithmic property that states \( \log_b(b^x) = x \ \). This property will be crucial for simplifying the given expression.
02

Apply the Logarithmic Property

Apply the property \( \log_b(b^x) = x \ \) to the given expression \( \log_q(q^{\sqrt{3}}) \ \). Here, the base \( q \) and the exponent \( \sqrt{3} \ \) fit the property exactly.
03

Simplify the Expression

According to the property, \( \log_q(q^{\sqrt{3}}) \ \) simplifies directly to \( \sqrt{3} \ \).

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.

Logarithmic Functions
Logarithmic functions are the inverses of exponential functions. If you have an exponential function of the form \(y = b^x\), the corresponding logarithmic form would be \(x = \log_b(y)\). This means that if you know the value of \(b^x\), you can determine \(x\) using the logarithmic function.
Exponential Expressions
An exponential expression involves a base raised to a power, like \(b^x\). In our exercise, we have the expression \(q^{\sqrt{3}}\). This indicates that the base \(q\) is raised to the exponent \(\sqrt{3}\). Such expressions are common in various fields like science, finance, and engineering.
Understanding how to work with exponential expressions is key when learning about logarithms because they are fundamentally related.
Properties of Logarithms
One of the most useful properties of logarithms is the power rule, which states \(\log_b(b^x) = x\). This property tells us that the logarithm of an exponential expression, where the base of the logarithm is the same as the base of the exponent, simplifies directly to the exponent. This is exactly the property used in our exercise to simplify \(\log_q(q^{\sqrt{3}})\) to \(\sqrt{3}\).
Other important logarithmic properties include:
  • Product Rule: \(\log_b(MN) = \log_b(M) + \log_b(N)\)
  • Quotient Rule: \(\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)\)
Mastering these properties will make working with logarithms much easier.

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

Use a graphing calculator to find the approximate solutions of the equation. $$4 \ln (x+3.4)=2.5$$

Use a graphing calculator to find the point \((s)\) of intersection of the graphs of each of the following pairs of equations. $$y=\left|1-3^{x}\right|, y=4+3^{-x^{2}}$$

Solve. $$e^{x}-2=-e^{-x}$$

Use a graphing calculator to find the approximate solutions of the equation. $$\log _{3} x+7=4-\log _{5} x$$

Solve using any method. $$5^{2 x}-3 \cdot 5^{x}+2=0$$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Discrete 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.