/*! 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 for \(x\) algebraically. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve for \(x\) algebraically. $$3^{x}=243$$

Short Answer

Expert verified
The solution for \(x\) in the equation \(3^x = 243\) is \(x = \frac{\log(243)}{\log(3)}.\)

Step by step solution

01

Apply Logarithm on Both Sides

Start by applying logarithm on both sides of the equation. It will be written as \(\log(3^x) = \log(243)\).
02

Use the Base Rule of Logarithms

By using the base rule of logarithms, the left side of the equation simplifies to \(x\log(3)\). The equation will now be written as \(x\log(3) = \log(243)\).
03

Solve for x

Since we want to solve for \(x\), we will divide both sides of the equation by \(\log(3)\). It will be written as \(x = \frac{\log(243)}{\log(3)}\).

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 91Ó°ÊÓ!

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. They can be of immense help when we need to solve equations where the variable is in an exponent, like in the equation \(3^{x}=243\). Logarithms have the basic form \(\log_b(a)\), where \(b\) is the base, \(a\) is the argument, and the result is the power to which we need to raise \(b\) to obtain \(a\). In other words, if \(b^y = a\), then \(\log_b(a) = y\).

When you're facing an exponential equation, and direct calculations are not feasible, switching to a logarithmic perspective can make the problem much more manageable. This is because logarithms can transform multiplication processes into addition, division into subtraction, and exponentiation into multiplication, simplifying the algebra involved.
Base Rule of Logarithms
The base rule of logarithms, also known as the change of base formula, is an essential tool for solving equations with exponents. This rule states that for any positive numbers \(a\), \(b\), and \(c\) (with \(b\) and \(c\) being different from 1), the logarithm of \(a\) with base \(b\) can be expressed in terms of logarithms with a different base \(c\) using this formula: \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\).

This formula allows you to convert a logarithm from one base to another, which is particularly helpful when using a scientific calculator, as they typically only have keys for logarithms in base 10 (common logarithms) or base \(e\) (natural logarithms). By deploying this technique, you can solve for variables within the logarithmic function, regardless of the base.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying algebraic expressions to isolate a variable or to put the expression into a more workable form. Techniques such as distribution, factoring, and employing arithmetic operations are fundamental. For example, when solving \(3^{x}=243\), you can perform algebraic manipulation to isolate \(x\).

Other instances include simplifying fractions, combining like terms, and manipulating equations to make use of properties like the logarithmic base rule. It's critical to understand the order of operations (PEMDAS/BODMAS) and to apply the correct steps systematically to work toward the solution of an equation.
Exponential to Logarithmic Form
Converting an exponential equation to logarithmic form is a key strategy in solving for variables trapped in exponents. The process relies on the understanding that if you have an equation in the form \(b^x = a\), you can rewrite it as \(x = \log_b(a)\). This conversion is crucial because it allows us to apply logarithmic operations to solve for \(x\), which can be more straightforward than tackling the exponent directly.

By using this method, the unwieldy exponential equation becomes a manageable logarithmic one. In the context of our example, \(3^{x} = 243\) was converted to \(x = \frac{\log(243)}{\log(3)}\) using this principle, enabling us to find \(x\) using arithmetic operations instead of exponential ones.

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

Verify that the hyperbolic cosine function \(\cosh (x)=\frac{e^{x}+e^{-x}}{2}\) is an even function.

Lead shielding is used to contain radiation. The percentage of a certain radiation that can penetrate \(x\) millimeters of lead shielding is given by \(I(x)=100 e^{-1.5 x}\) a. What percentage of radiation, to the nearest tenth of a percent, will penetrate a lead shield that is 1 millimeter thick? b. How many millimeters of lead shielding are required so that less than \(0.05 \%\) of the radiation penetrates the shielding? Round to the nearest millimeter.

If we assume that air resistance is proportional to the square of the velocity, then the velocity \(v\) in feet per second of an object \(t\) seconds after it has been dropped is given by $$v=50\left(\frac{e^{1.6 t}-1}{e^{1.6 t}+1}\right)$$ a. In how many seconds will the velocity be 20 feet per second? b. Determine the horizontal asymptote for the graph of this function. c. Write a sentence that describes the meaning of the horizontal asymptote in the context of this problem.

The number of bass in a lake is given by $$ P(t)=\frac{3600}{1+7 e^{-0.05 t}} $$ -where \(t\) is the number of months that have passed since the lake was stocked with bass. a. How many bass were in the lake immediately after it was stocked? b. How many bass were in the lake 1 year after the lake was stocked? c. What will happen to the bass population as \(t\) increases without bound?

Make use of the factorial function, which is defined as follows. For whole numbers \(n\), the number \(n !\) (which is read "n factorial") is given by $$n !=\left\\{\begin{array}{ll} n(n-1)(n-2) \cdots 1, & \text { if } n \geq 1 \\\1, & \text { if } n=0 \end{array}\right.$$Thus, \(0 !=1\) and \(4 !=4 \cdot 3 \cdot 2 \cdot 1=24\) STIRLING'S FORMULA Stirling's Formula (after James Stir\(\operatorname{lin} g, 1692-1770)\) 1$$n !=\left(\frac{n}{e}\right)^{n} \sqrt{2 \pi n}$$ is often used to approximate very large factorials. Use Stirling's Formula to approximate \(10 !\), and then compute the ratio of Stirling's approximation of \(10 !\) divided by the actual value of \(10 !\), which is 3,628,800

See all solutions

Recommended explanations on Math Textbooks

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.