/*! 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 2 Solve the exponential equations ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 2

Solve the exponential equations exactly for \(x\). $$169^{x}=13$$

Short Answer

Expert verified
The solution is \(x = \frac{1}{2}\).

Step by step solution

01

Recognize the Relationship Between the Bases

The equation is \(169^x = 13\). Notice that \(169\) can be expressed as \(13^2\). This is because \(169 = 13 \times 13\). This is the key step to solving the equation as we can rewrite the equation with a common base.
02

Rewrite the Equation Using the Common Base

Substitute \(169\) with \(13^2\) in the original equation:\[(13^2)^x = 13\]This simplifies to \(13^{2x} = 13^1\) since raising a power to a power multiplies the exponents.
03

Set the Exponents Equal

Since the bases \(13^{2x}\) and \(13^1\) are the same, we can set the exponents equal to each other:\[2x = 1\]
04

Solve for x

Solve the equation \(2x = 1\) by dividing both sides by 2:\[x = \frac{1}{2}\]This gives the exact solution for \(x\).

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.

Exponentiation
Exponentiation is a mathematical operation where a number, known as the base, is raised to a certain power, which is called the exponent. When we write something like \(b^n\), we are saying that the base \(b\) is multiplied by itself \(n\) times. This makes exponential form a convenient way to express very large or very small numbers. For example:
  • \(2^3 = 2 \times 2 \times 2 = 8\)
  • \(5^4 = 5 \times 5 \times 5 \times 5 = 625\)
Exponentiation has several key properties:
  • The base raised to the power of zero is always 1: \(b^0 = 1\) for any non-zero \(b\).
  • A base raised to the power of one is the base itself: \(b^1 = b\).
  • When you multiply powers with the same base, you add the exponents: \(b^m \times b^n = b^{m+n}\).
  • When you raise a power to another power, you multiply the exponents: \((b^m)^n = b^{m \cdot n}\).
Understanding these properties helps to simplify complex expressions and solve exponential equations effectively.
Common Base
The concept of a common base is crucial in solving exponential equations, like the problem \(169^x = 13\). The idea is to express all terms in the equation with the same base, if possible. Let's see how this helps identify the solution:
  • First, recognize the relationship between the numbers. Notice that 169 is actually \(13^2\), which gives us a common base of 13 for both sides of the equation.
  • Rewriting \(169^x\) as \((13^2)^x\) helps simplify the equation, turning it into \(13^{2x}\). Now, you deal with exponents more easily.
By rewriting terms using a common base, the equation becomes simpler and the exponents can be directly compared. This method is widely used in mathematical problems involving powers and roots, allowing you to avoid complex calculations and focus on the properties of exponents.
Solving Equations
Solving equations, especially exponential ones, requires a clear strategy. Consider the exponential equation \(169^x = 13\). Similar to other equations, the goal is to isolate \(x\).

Here's the step-by-step approach:
  • Recognize any shared bases and express all terms with this base, as was done when identifying \(169 = 13^2\).
  • Set the exponents equal once you have the same base on both sides of the equation. This works because if \(b^m = b^n\), then \(m = n\).
  • Solve the simpler resulting equation for \(x\). For \(13^{2x} = 13^1\), you set \(2x = 1\).
  • Finally, solve for \(x\) by dividing both sides by the necessary coefficient. Here, you divide by 2 to find \(x = \frac{1}{2}\).
Following these steps makes solving exponential equations less daunting. Understanding the significance of each step and operation ensures you approach similar problems with confidence.

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

In calculus we prove that the derivative of \(f+g\) is \(f^{\prime}+g^{\prime}\) and that the derivative of \(f-g\) is \(f^{\prime}-g^{\prime} .\) It is also shown in calculus that if \(f(x)=\ln x\) then \(f^{\prime}(x)=\frac{1}{x}\) Use these properties to find the derivative of \(f(x)=\ln \frac{1}{x}\)

Refer to the following: In calculus, we find the derivative, \(f^{\prime}(x),\) of a function \(f(x)\) by allowing \(h\) to approach 0 in the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of functions involving exponential functions. Find the difference quotient of the exponential growth model \(f(x)=P e^{k x},\) where \(P\) and \(k\) are positive constants.

Suppose that both logistic growth models \(f(t)=\frac{a_{1}}{1+c_{1} e^{-k_{i} t}}\) and \(g(t)=\frac{a_{2}}{1+c_{2} e^{-k_{s} t}}\) have horizontal asymptote \(y=100\) What can you say about the corresponding carrying capacities?

Solve \(y=\frac{3000}{1+2 e^{-0.2 t}}\) for \(t\) in terms of \(y\).

The hyperbolic tangent is defined by tanh \(x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\) Find its inverse function \(\tanh ^{-1} x\).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

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.