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Solve the exponential equation algebraically. Approximate the result to three decimal places. $$2^{x}=3^{x+1}$$

Short Answer

Expert verified
The solution to the equation \( 2^{x} = 3^{x+1} \) approximated to three decimal places is calculated as per the steps above. Refer to Step 5 for the exact value, rounded to three decimal places.

Step by step solution

01

Apply logarithm to both sides of the equation

We will apply the natural logarithm (ln) to both sides of the equation, which will turn into:\( \ln(2^{x}) = \ln(3^{x+1}) \)
02

Use the property of logarithms

Using the property of logarithms that allows to bring down exponents in front of the logarithm:\( x \cdot \ln(2) = (x+1) \cdot \ln(3) \)
03

Distribute and Rearrange

Next the logarithm expression on the right \( \ln(3)*(x+1) \) is expanded and then rearranged to isolate x terms on one side and constant term on the other side:\( x \cdot \ln(2) = x \cdot \ln(3) + \ln(3) \) becomes \( x \cdot \ln(2) - x \cdot \ln(3) = \ln(3) \)
04

Solving for x

Now the x terms on the left side can be factored out, and finally by dividing both sides by the remaining constant on the left, x can be isolated:\( x = \frac{\ln(3)}{\ln(2) - \ln(3)} \)
05

Approximate the value of x

By substituting the known values of \( \ln(2) \) and \( \ln(3) \), we get an approximated value of x rounded to three decimal places. Remember to substitute these values in a calculator with the set decimal approximation.

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Key Concepts

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

Logarithmic Properties
Logarithmic properties are essential rules for simplifying and solving logarithmic expressions. When solving exponential equations like 2^{x} = 3^{x+1}, applying a logarithm to both sides is a crucial step. This approach is grounded in one of the key logarithmic properties: the power rule, which states that log_b(a^n) = n*log_b(a). This allows the exponent to be brought down in front, making the equation easier to handle. For instance, (2^{x}) becomes x*(2), and similarly with (3^{x+1}). This property is incredibly helpful in isolating the variable x when it appears as an exponent, which is often the case in exponential equations.

Understanding and applying logarithmic properties correctly can transform complex exponential equations into more manageable algebraic ones, paving the way for finding solutions using familiar techniques.
Algebraic Manipulation
Algebraic manipulation involves rearranging equations and expressions to solve for unknown values or to simplify them. After applying logarithms to both sides of an exponential equation, we often get terms that require algebraic skills to manipulate. Distributing, combining like terms, and isolating variables are key steps. For example, from the equation x (2) = (x+1) (3), expanding and rearranging terms gives x (2) - x (3) = (3). This demonstrates the importance of being adept with algebraic manipulation.

It enables the clear isolation of terms containing the variable we're solving for, allowing us to use addition, subtraction, multiplication, or division to reach the solution. Mastering these techniques is fundamental in extracting the variable from the complex interplay of logarithmic operations.
Natural Logarithm
The natural logarithm, represented as , is a logarithm to the base e, where e is an irrational and transcendental number approximately equal to 2.71828. In the context of solving exponential equations, using the natural logarithm can be particularly useful due to its unique properties. For instance, (e^x) = x and e^((x)) = x, often called the inverse property of logarithms and exponentials, make a natural choice for dealing with exponential expressions.

The natural logarithm arises in various domains, such as compound interest in finance, population growth in biology, and radioactive decay in physics, which hints at its fundamental significance and utility in mathematics. When dealing with exponential equations, particularly when the base is e, the natural logarithm simplifies calculations and makes the process of finding solutions more straightforward.
Exponentiation
Exponentiation is a mathematical operation involving a base raised to an exponent, encapsulating the concept of repeated multiplication. In the equation 2^{x} = 3^{x+1}, both sides embody exponentiation with different bases. Solving exponential equations typically requires transforming the equation so that the bases are comparable or eliminated, hence the use of logarithms.

Exponentiation is prevalent in many areas of mathematics and real-life applications, from computing powers in programming algorithms to representing exponential growth or decay in scientific models. Developing a sound understanding of how to manipulate exponential expressions is crucial, as it's this very manipulation—often with logarithms—that allows us to solve for the exponent and interpret these expressions in practical scenarios.

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Most popular questions from this chapter

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x \ln \left(\frac{1}{x}\right)-x=0$$

Condense the expression to the logarithm of a single quantity. $$2 \ln 8+5 \ln (z-4)$$

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log _{4} x-\log _{4}(x-1)=\frac{1}{2}$$

Determine whether the statement is true or false given that \(f(x)=\ln x .\) Justify your answer. $$f(a x)=f(a)+f(x), \quad a>0, \quad x>0$$

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). $$\begin{aligned} &\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)\\\ &\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right) \end{aligned}$$ (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 resulting 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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