/*! 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 15 Solve the exponential equation. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the exponential equation. Round to three decimal places, when needed. $$3\left(1.3^{x}\right)=5$$

Short Answer

Expert verified
After plugging these values into a calculator, the solution for \(x\) is approximately 0.630.

Step by step solution

01

Isolate the Exponential Term

First, divide both sides of the equation by 3: \(\frac{3(1.3^{x})}{3} = \frac{5}{3}\) which simplifies to: \(1.3^{x} = \frac{5}{3}\).
02

Apply the Logarithm

Now, apply the natural logarithm, \(ln\), to both sides of the equation: \(ln(1.3^{x}) = ln\left(\frac{5}{3}\right)\). The purpose of this step is to 'bring down' the exponent \(x\).
03

Use the Logarithm Property

Apply the property of logarithms that states \(ln(a^{p}) = p * ln(a)\), where \(a = 1.3\), and \(p = x\). The equation becomes: \(x*ln(1.3) = ln\left(\frac{5}{3}\right)\).
04

Solve for \(x\)

Now, isolate \(x\) by dividing both sides of the equation by \(ln(1.3)\) to get the solution: \(x = \frac{ln\left(\frac{5}{3}\right)}{ln(1.3)}\).

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Key Concepts

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

Exponential Equation
An exponential equation is a type of equation where a constant base is raised to a variable exponent. In the given exercise, the equation is in the form of \( base^{exponent} = value \), with the base being 1.3 and the exponent being the variable \( x \). To solve it, one must isolate the variable, which involves a series of algebraic manipulations.

Understanding the behavior of exponential functions is crucial in solving such equations. For instance, exponential growth occurs when the base is greater than one, as is the case with 1.3 in our exercise. This nature is important because it informs us that as \( x \) increases, the value of \( 1.3^x \) grows exponentially.
Natural Logarithm
The natural logarithm, written as \( \ln \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. We often employ the natural logarithm to solve exponential equations because it provides the inverse operation that allows us to 'undo' the exponential function.

By applying \( \ln \) to both sides of an exponential equation, as done in the second step of our problem, we leverage the inherent relationship between exponents and logarithms to simplify the equation and make the variable \( x \) accessible for isolation and solution.
Logarithm Properties
Logarithms come with several properties that make them incredibly useful for solving equations. One crucial property is the power rule, which states that \( \ln(a^p) = p \ln(a) \). This allows us to move the exponent of the logarithmic argument out in front as a multiplier.

In the given exercise, by applying this property in step 3, we transform the problem from its original exponential form into a linear equation where \( x \) can be easily isolated. Other logarithm properties include the product rule, the quotient rule, and the change of base formula—which can all be instrumental in solving different types of logarithmic and exponential equations.
Isolate Variable
Isolating the variable is the pivotal step in solving any algebraic equation. It generally involves rearranging the equation so that the variable of interest is on one side of the equation and everything else is on the other side.

In the context of exponential equations, once we've used logarithms to bring down the exponent, as in step 3 of our solution, the next step is to isolate \( x \). This is generally done by performing basic algebraic operations such as addition, subtraction, multiplication, or division. In our specific problem, after applying the natural logarithm properties, isolating \( x \) became a matter of dividing both sides of the equation by \( \ln(1.3) \), successfully obtaining the value of \( x \) in step 4.

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Most popular questions from this chapter

Solve the logarithmic equation and eliminate any extraneous solutions. If there are no solutions, so state. $$\log _{2}(x+5)=\log _{2}(x)+\log _{2}(x-3)$$

The following data gives the percentage of women who smoked during pregnancy for selected years from 1994 to \(2002 .\) (Sournce: National Center for Health Statistics) $$\begin{array}{|c|c|} \hline\text { Year } & \text { Percent Smoking } \\\\\text { Yuring Pregnancy } \\ \hline1994 & 14.6 \\\1996 & 13.6 \\\1998 & 12.9 \\\2000 & 12.2 \\\2001 & 12.0 \\\2002 & 11.4\\\\\hline\end{array}$$ (a) From examining the table, what is the general relationship between the year and the percentage of women smoking during pregnancy? (b) Let \(t\) be the number of years after \(1993 .\) Here, \(t\) starts at 1 because in 0 is undefined. Make a scatter plot of the data and find the natural logarithmic function of the form \(p(t)=a \ln t+b\) that best fits the data. Why must a be negative? (c) Project the percentage of women who will smoke during pregnancy in the year 2007.

Evaluate the expression to four decimal places using a calculator. $$\ln \sqrt{2}$$

Determine how long it takes for the given investment to double if \(r\) is the interest rate and the interest is compounded continuously. Assume that no withdrawals or further deposits are made. Initial amount: \(\$ 1500 ; r=6 \%\)

The value of a 2006 S-type Jaguar is given by the function $$v(t)=43,173(0.8)^{t}$$ where \(t\) is the number of years since its purchase and \(v(t)\) is its value in dollars. (Source: Kelley Blue Book) (a) What was the Jaguar's initial purchase price? (b) What percentage of its value does the Jaguar S-type lose each year? (c) How many years will it take for the Jaguar S-type to reach a value of \(\$ 22,227 ?\)

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