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Chapter 4: Problem 27

Solve the exponential equation. $$\left(\frac{1}{8}\right)^{x}=64$$

Short Answer

Expert verified
The solution to the equation \(\left(\frac{1}{8}\right)^{x}=64\) is \(x = -2\).

Step by step solution

01

Rewrite both sides of the equation with base 2

Write \(\left(\frac{1}{8}\right) = 2^{-3}\) and \(64 = 2^{6}\). This leads us to \(2^{-3x} = 2^{6}\).
02

Equate exponents

Since the bases are equal (\(2\) in this case), it implies that the exponents are also equal. This leads to \(-3x = 6\).
03

Solve for x

Solve for \(x\), which gives you \(x = \frac{-6}{-3} = -2\).

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Key Concepts

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

Base Conversion
When dealing with exponential equations, it is often crucial to work with similar or the same bases. This involves converting the given numbers into expressions with the common base. In our exercise, we began with the equation \(\left(\frac{1}{8}\right)^{x}=64\). Both these numbers can be written as powers of 2:
  • The fraction \(\frac{1}{8}\) converts to \(2^{-3}\) because \(8 = 2^3\) and \(\frac{1}{8} = 8^{-1}\), which simplifies to \(2^{-3}\).
  • The number 64 is equal to \(2^6\) since \(64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\).
Converting numbers this way allows you to compare exponents directly, a strategy often needed in more advanced exponential problems. It's crucial to choose a base that simplifies the equation, often the smallest prime base common between both numbers.
Equating Exponents
Once the bases of the two sides of the equation are the same, we can move to the next step: equating the exponents. In the equation \(2^{-3x} = 2^{6}\), the bases are already equal, which directly allows us to set their exponents equal to each other.

This simplifies the equation to solving \(-3x = 6\). This principle is founded on the property of exponential functions, which states that if \(a^m = a^n\), then it must be the case that \(m = n\).

Therefore, solving the exponent gives us the value for \(x\). Keep in mind that this ability to equate exponents only works when the bases are identical, which is why base conversion is such an important preliminary step.
Solving for x
After equating the exponents in our exponential equation, the next task is solving for \(x\). From \(-3x = 6\), solving for \(x\) involves isolating it on one side of the equation. We can do this by dividing both sides by \(-3\).
  • First, divide both sides of the equation by \(-3\): \(-3x = 6\) becomes \(x = \frac{6}{-3}\).
  • After simplifying, this gives \(x = -2\).
This process requires using basic algebraic principles, where you perform equal operations on both sides of an equation to maintain its balance. Solving for \(x\) in such cases not only tests your algebra skills but also reinforces your understanding of exponential equations and their properties.

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Most popular questions from this chapter

(a) complete the table to find an interval containing the solution of the equation, (b) use a graphing utility to graph both sides of the equation to estimate the solution, and (c) solve the equation algebraically. Round your results to three decimal places. $$6 \log _{3}(0.5 x)=11$$ $$\begin{array}{|l|l|l|l|l|l|}\hline x & 12 & 13 & 14 & 15 & 16 \\\\\hline 6 \log _{3}(0.5 x) & & & & & \\\\\hline\end{array}$$

Evaluate the function for \(f(x)=3 x+2\) and \(g(x)=x^{3}-1.\) $$(f+g)(2)$$

Use the zero or root feature of a graphing utility to approximate the solution of the logarithmic equation. $$\log _{10} x+e^{0.5 x}=6$$

Use a graphing utility to create a scatter plot of the data. Decide whether the data could best be modeled by a linear model, an exponential model, or a logarithmic model. $$(1,2.0),(1.5,3.5),(2,4.0),(4,5.8),(6,7.0),(8,7.8)$$

Use the regression feature of a graphing utility to find a power model \(y=a x^{b}\) for the data and identify the coefficient of determination. Use the graphing utility to plot the data and graph the model in the same viewing window. $$(0.5, 1.0), (2, 12.5), (4, 33.2), (6, 65.7), (8, 98.5),(10, 150.0)$$
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