/*! 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 57 Find the logarithm by applying t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the logarithm by applying the definition of logarithm $$x=\log _{2} 32$$

Short Answer

Expert verified
\(x = 5\)

Step by step solution

01

Understand the Definition of a Logarithm

The logarithm of a number is the power to which the base must be raised to produce that number. In other words, if we have \(x = \log_{b}(y)\), it means that \(b^x = y\).
02

Apply the Definition to the Given Problem

We need to find \(x\) such that \(2^x = 32\). This means we are looking for the exponent \(x\) to which the number 2 must be raised to get 32.
03

Solve for \(x\)

Since 32 is a power of 2 (specifically, \(2^5\)), we can equate \(2^x\) to \(2^5\) because \(32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^5\). Therefore, \(x = 5\).

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 91Ó°ÊÓ!

Key Concepts

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

Logarithm Calculation
Logarithm calculation often stirs confusion, but it's simply a way to unravel the power or exponent that a given base must be raised to, to arrive at a specific value. Let's consider our example, where you are tasked to find the value of x in the equation x = \(\log_{2}{32}\). This is asking, \

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.