/*! 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 24 Solve the exponential equation a... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the exponential equation algebraically. Approximate the result to three decimal places. \(4^{-3 t}=0.10\)

Short Answer

Expert verified
The solution to the given equation is \(t \approx 0.482\) when rounded to three decimal places.

Step by step solution

01

Write the equation

First write down the given equation, which is \(4^{-3 t}=0.10\).
02

Applying natural logarithm

To eliminate the exponential expression, apply a natural logarithm on both sides of the equation. This gives \(\ln(4^{-3t})=\ln(0.10)\).
03

Applying logarithm properties

Next, apply logarithm properties to simplify the left side of the equation. The exponent \(-3t\) can be brought out front using the rule \(\ln(a^b) = b \ln(a)\), providing \(-3t \ln(4) = \ln(0.10)\).
04

Isolate the variable \(t\)

The final step is to isolate the variable. Dividing both sides by \(-3 \ln(4)\) will isolate \(t\). This gives the solution as \(t = \frac{\ln(0.10)}{-3 \ln(4)}\).
05

Calculating the final result

The calculation should be performed to determine the numerical value of \(t\), and it should be rounded to three decimal places. Inputting the numbers and performing the calculation yield \(t \approx 0.482\).

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Key Concepts

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

Understanding Natural Logarithms
When solving exponential equations, one of the most effective tools is the natural logarithm, denoted as \(\ln\). This special function is actually a specific type of logarithm that uses the number \(e\) (approximately 2.71828) as its base. It's 'natural' because the number \(e\) arises naturally in many areas of mathematics, particularly when dealing with continuous growth or decay processes.

Natural logarithms turn the process of multiplication and exponentiation into addition and multiplication, respectively. This property is particularly useful when dealing with exponential equations because it allows us to bring down exponents for easier handling. For instance, in our equation \(4^{-3t} = 0.10\), to solve for \(t\), we take the natural logarithm of both sides, which transforms the problem into an algebraic one that we can tackle using more familiar operations.
Leveraging Logarithm Properties
Logarithm properties are rules that allow us to manipulate logarithmic expressions and make them easier to manage. These properties can simplify complex logarithmic equations into forms that can be easily solved. A foundational property is the \emph{power rule} which states that \(\ln(a^b) = b \ln(a)\).

In the context of our exercise, this rule is applied to move the exponent to the front, turning \(\ln(4^{-3t})\) into \(\line{-3t \ln(4)}\). Other useful properties include the \emph{product rule}, \emph{quotient rule}, and \emph{logarithm of one rule}. Understanding and correctly applying these logarithm properties is critical when trying to isolate a variable within an exponential function.
Isolating the Variable
To find the solution to an equation, we often need to \emph{isolate the variable}—that is, move all instances of the variable to one side of the equation, leaving a numerical expression on the other side. Doing so turns the abstract equation into a solvable problem where the value of the variable becomes clear.

In our exercise, we isolated \(t\) by dividing both sides of the equation by the numerical coefficient \(\line{-3 \ln(4)}\) that was attached to it. The resulting expression after isolation allows us to directly find the value of \(t\) using simple arithmetic operations. Mastering the skill of variable isolation is essential, especially in algebra where variables are ubiquitous and are the primary focus of most problems.
Algebra with Exponential Functions
Exponential function algebra involves a series of steps that are used to solve equations where the variable is part of the exponent—hence the name exponential equations. It typically requires the understanding of exponents and experience with algebraic manipulation.

These functions are characterized by their base raised to a variable exponent, as seen in \(4^{-3t}\). They often describe situations of rapid growth or decay and occur naturally in many scientific contexts, like population growth, radioactive decay, and compound interest. Solving exponential equations algebraically often involves taking logarithms of both sides, utilizing logarithm properties to bring down the exponent, and rearranging the terms to isolate the variable. This allows us to move from the abstract representation of a phenomenon to a quantifiable and specific value, as was required in determining \(t\) for our equation \(4^{-3t} = 0.10\).

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Most popular questions from this chapter

Using Properties of Logarithms In Exercises \(15-20\) , use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\log _{4} 8$$

Think About It For how many integers between 1 and 20 can you approximate natural logarithms, given the values \(\ln 2 \approx 0.6931,\) ln \(3 \approx 1.0986,\) and ln 5\(\approx 1.6094 ?\) Approximate these logarithms (do not use a calculator).

True or False? In Exercises \(97-102,\) determine whether the statement is true or false given that \(f(x)=\ln x .\) Justify your answer. $$f(a x)=f(a)+f(x), \quad a>0, \quad x>0$$

Using Properties of Logarithms In Exercises \(15-20\) , use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\ln \frac{6}{e^{2}}$$

Comparing Models A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C} .\) The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$ \begin{array}{l}{\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)} \\\ {\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right)}\end{array} $$ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\) . (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This rasulting line has the form \(\ln (T-21)=a t+b\) . Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$ \left(t, \frac{1}{T-21}\right) $$ Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$ \frac{1}{T-21}=a t+b $$ Solve for \(T,\) and use the graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

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