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Chapter 11: Problem 32

Solve each equation. $$4^{3 x}=\left(\frac{1}{2}\right)^{1-x}$$$

Short Answer

Expert verified
The solution is \(x = -0.2\).

Step by step solution

01

Rewrite the Equation with a Common Base

First, recognize that both the left and right sides of the equation can be written with a common base. Notice that \(4 = 2^2\) and \(\frac{1}{2} = 2^{-1}\). Rewrite the equation accordingly:\[4^{3x} = \left(\frac{1}{2}\right)^{1-x} \rightarrow (2^2)^{3x} = (2^{-1})^{1-x}\]
02

Simplify the Exponents

Apply the power rule \((a^m)^n = a^{mn}\) to both sides of the equation:\[2^{6x} = 2^{-(1-x)}\]
03

Set the Exponents Equal

Since the bases are the same, set the exponents equal to each other:\[6x = -(1 - x)\]
04

Solve for x

Solve the equation for \(x\):\[6x = -1 + x \rightarrow 6x - x = -1 \rightarrow 5x = -1 \rightarrow x = \frac{-1}{5} = -0.2\]

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Key Concepts

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

Common Base
When solving exponential equations, it's often helpful to rewrite the equation using a common base. In this problem, we have the equation \(4^{3x} = \frac{1}{2}^{1-x}\). Notice that the number 4 can be expressed as \(2^2\) and the fraction \(\frac{1}{2}\) can be written as \(2^{-1}\). By rewriting both sides with the common base of 2, the equation becomes easier to manage. This step transforms our equation into \((2^2)^{3x} = (2^{-1})^{1-x}\). Using a common base simplifies the comparison of the exponents.
Exponents
Exponents are the numbers that indicate how many times a base is multiplied by itself. In the equation \((2^2)^{3x} = (2^{-1})^{1-x}\), we use properties of exponents to simplify each side. Specifically, we use the power rule, \((a^m)^n = a^{mn}\).
Applying this rule, the left side simplifies to \(2^{6x}\) and the right side to \(2^{-(1-x)}\). Now we have a new equation: \(2^{6x} = 2^{-(1-x)}\). At this stage, solving the equation becomes more straightforward since the bases are now identical.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying mathematical expressions. With our new equation \(2^{6x} = 2^{-(1-x)}\), the next step is to set the exponents equal to each other. This is because if \(a^m = a^n\), then it must follow that \(m = n\).
Therefore, we set:\(6x = -(1-x)\). To solve for \(x\), combine like terms and isolate \(x\):
- Add \(x\) to both sides: \(6x + x = -1\)
- Combine: \(7x = -1\)
- Finally, divide by 7: \(x = \frac{-1}{5}\).
Equation Solving
Equation solving is the process of finding the value of the variable that makes the equation true. In our case, we started with \(4^{3x} = \frac{1}{2}^{1-x}\) and through a series of steps鈥攔ewriting in common bases, simplifying exponents, and algebraic manipulation鈥攚e've narrowed it down to the simpler equation \(7x = -1\). To isolate \(x\), we performed basic arithmetic operations:
. Combined like terms:
- Divided both sides by 7:
This gives us the solution \(x = \frac{-1}{5} = -0.2\). At each step, ensuring the accuracy of our manipulations was key to finding the correct value of \(x\).

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Most popular questions from this chapter

Determine whether each equation is true or false. $$ \ln (25)=2 \cdot \ln (5) $$

Use the base-change formula to find each logarithm to four decimal places. $$\log _{1 / 3}(3.5)$$

Fill in the missing entries in each table. $$\begin{array}{c|c|c|c|c|c}x & \frac{1}{125} & & 1 & & 625 \\\\\hline \log _{5}(x) & & & -2 & & 1 &\end{array}$$

Use a calculator to evaluate each logarithm. Round answers to four decimal places. $$\ln (0.23)$$

The U.S.G.S. measures the quality of a water sample by using the diversity index \(d\), given by \(d=-\left[p_{1} \cdot \log _{2}\left(p_{1}\right)+p_{2} \cdot \log _{2}\left(p_{2}\right)+\cdots+p_{n} \cdot \log _{2}\left(p_{n}\right)\right]\), where \(n\) is the number of different taxons (biological classifications) represented in the sample and \(p_{1}\) through \(p_{n}\) are the percentages of organisms in each of the \(n\) taxons. The value of \(d\) ranges from 0 when all organisms in the water sample are the same to some positive number when all organisms in the sample are different. If two- thirds of the organisms in a water sample are in one taxon and one-third of the organisms are in a second taxon, then \(n=2\) and $$ d=-\left[\frac{2}{3} \log _{2}\left(\frac{2}{3}\right)+\frac{1}{3} \log _{2}\left(\frac{1}{3}\right)\right] $$ Use the properties of logarithms to write the expression on the right-hand side as \(\log _{2}\left(\frac{3 \sqrt{2}}{2}\right) .\) (In Section 11.4 you will learn how to evaluate a base-2 logarithm using a calculator.)
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