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Chapter 8: Problem 116

Solve: \(2^{|2 x-3|}=64\)

Short Answer

Expert verified
The solutions are \(x = \frac{9}{2}\) and \(x = -\frac{3}{2}\).

Step by step solution

01

Recognize the Exponential Form

First, identify the common base in the equation. The equation given is \(2^{|2x-3|} = 64\). Recognize that 64 is a power of 2, specifically \(64 = 2^6\). This allows us to rewrite the equation as \(2^{|2x-3|} = 2^6\).
02

Equate the Exponents

Since the bases are the same, we can equate the exponents. This gives us \(|2x-3| = 6\).
03

Solve for the Absolute Value

Recall the definition of an absolute value: \(|a| = b\) implies \(a = b\) or \(a = -b\). Thus, from \(|2x-3| = 6\), we create two equations: \(2x-3 = 6\) and \(2x-3 = -6\).
04

Solve the First Equation

Let's solve \(2x-3 = 6\).Add 3 to both sides:\[2x - 3 + 3 = 6 + 3\]\[2x = 9\]Divide both sides by 2:\[x = \frac{9}{2}\]
05

Solve the Second Equation

Now, solve \(2x-3 = -6\).Add 3 to both sides:\[2x - 3 + 3 = -6 + 3\]\[2x = -3\]Divide both sides by 2:\[x = -\frac{3}{2}\]
06

Write the Final Solution

The solutions to the equation are \(x = \frac{9}{2}\) and \(x = -\frac{3}{2}\). Both solutions satisfy the original equation.

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Key Concepts

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

Understanding Absolute Value
The absolute value of a number represents its distance from zero on a number line, regardless of its direction. It is always non-negative. For example, the absolute value of both 3 and -3 is 3. This concept is noted with vertical bars:
  • For a positive number, the absolute value is the number itself, e.g., \(|5| = 5\).
  • For a negative number, the absolute value is the positive version of that number, e.g., \(|-7| = 7\).
  • For zero, the absolute value is \(0\), as zero is neither positive nor negative: \(|0| = 0\).
In equations, the absolute value can introduce two possible solutions. This is because if \(|a| = b\), then the expression inside the absolute value bars, \(a\), could equal \(b\) or \(-b\). Therefore, absolute value equations can be split into two separate equations to solve.
Steps in Equation Solving
Equation solving is all about finding the value(s) of the variable that make the equation true. To do this, follow a logical sequence of steps:
  • First, simplify both sides of the equation if needed, combining like terms or eliminating parentheses.
  • Move variable terms to one side and constants to the other by adding, subtracting, multiplying, or dividing both sides by the same number.
  • Once simplified, solve for the variable by isolating it. For instance, if you have an equation like \(2x = 8\), you isolate \(x\) by dividing both sides by 2, resulting in \(x = 4\).
In our original problem, after equating the bases, we had \(|2x-3| = 6\). From there, it was split into two scenarios to explore both possible values: \(2x-3 = 6\) and \(2x-3 = -6\), which were then solved step-by-step.
Exploring Powers of Numbers
When we talk about powers of numbers, we are referring to the repeated multiplication of a number by itself. In simpler terms, when a number is raised to an exponent, it is multiplied by itself that many times. For example:
  • \(2^3\) means \(2\) is multiplied by itself twice: \(2 \times 2 \times 2 = 8\).
  • \(3^2\) means \(3\) times itself once: \(3 \times 3 = 9\).
Recognizing powers is a useful skill when solving exponential equations, as it allows you to equate exponents directly if the bases are the same. In our original equation, \(2^{|2x-3|} = 64\), identifying that \(64 = 2^6\) allowed us to simplify the equation and focus on the exponents. This equivalence occurs because exponential functions are one-to-one, meaning a unique exponent corresponds to a given power when the bases are the same.

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Most popular questions from this chapter

In the following problems, simplify each expression by performing the indicated operations and solve each equation. $$\frac{4}{m^{2}-9}+\frac{5}{m^{2}-m-12}=\frac{7}{m^{2}-7 m+12}$$

In the following problems, simplify each expression by performing the indicated operations and solve each equation. $$\left(2 x^{2}-15 x+25\right) \div \frac{2 x^{2}-3 x-5}{x+1}$$

Solve each equation. If the equation is an identity or a contradiction, so indicate. $$ \frac{3}{2}(a-4)=2(a-3)-\frac{a}{2} $$

Write some comments to the student who wrote the following solution, explaining his misunderstanding. Multiply: $$\begin{aligned}\frac{1}{x} \cdot \frac{3}{2} &=\frac{1}{x} \cdot \frac{2}{2} \cdot \frac{3}{2} \cdot \frac{x}{x} \\\&=\frac{2}{2 x} \cdot \frac{3 x}{2 x} \\\&=\frac{6 x}{4 x^{2}}\end{aligned}$$

In the following problems, simplify each expression by performing the indicated operations and solve each equation. $$\frac{3}{s-2}+\frac{s-14}{2 s^{2}-3 s-2}-\frac{4}{2 s+1}=0$$
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