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Chapter 3: Problem 15

Solve each exponential equation in Exercises \(1-22\) by expressing each side as a power of the same base and then equating exponents $$6^{\frac{x-3}{4}}=\sqrt{6}$$

Short Answer

Expert verified
The solution to the exercise is \(x = 5\).

Step by step solution

01

Express both sides as powers of the same base

Let's first express the right-hand side \(\sqrt{6}\) in a different form using exponentiation. We know that the square root of a number (n) is equivalent to raising that number (n) to the power of 0.5. So, we can rewrite \(\sqrt{6}\) as \(6^{0.5}\). Our equation then becomes: \(6^{\frac{x-3}{4}} = 6^{0.5}\)
02

Equate the exponents

Now that both sides of the equation are written as the same number (6) raised to a power, we can set the exponents equal to each other. So, we get the equation: \(\frac{x-3}{4} = 0.5\)
03

Solve for unknown 'x'

To solve for 'x' we multiply through by 4 to get rid of the fraction, which gives us: \(x-3 = 4*0.5\). Then, simplify the right hand side and add 3 to both sides to solve for 'x': \(x = 4*0.5 + 3\)

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Key Concepts

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

Expressing Powers
In exponential equations, expressing powers is a vital skill that can simplify complex equations. When both sides of an equation can be expressed as powers of the same base, solving becomes significantly easier. This is because exponents on a common base can be equated, transforming an exponential equation into a much simpler form. For instance, if given an equation like \(6^{\frac{x-3}{4}}=\sqrt{6}\), the first step would be to express \(\sqrt{6}\) using exponential notation. We identify that the square root of a number, such as 6, is equivalent to that number raised to the power of 0.5: \(\sqrt{6} = 6^{0.5}\). This allows us to rewrite the equation in a consistent and manageable form, \(6^{\frac{x-3}{4}} = 6^{0.5}\), setting the stage for the next steps.
Same Base
The concept of the "same base" is crucial when solving exponential equations. When both sides of an equation are represented as powers of the same base, you unlock the method of equating exponents. This simplifies the equation immensely. In practice, as demonstrated in the exercise, the equation \(6^{\frac{x-3}{4}} = 6^{0.5}\) can be simplified by acknowledging that both sides rely on the base 6. This alignment of bases means that the operations focus solely on the exponents, discarding the base and making the original equation less complex to solve. By equating the exponents, you translate an exponential equation into a linear or simpler form, which is often easier to handle and solve.
Solving Equations
Solving exponential equations typically involves expressing both sides with the same base and then focusing on the exponents. After achieving a common base, the next step is to equate the exponents. For example, with the equation \(6^{\frac{x-3}{4}} = 6^{0.5}\) after expressing both as powers of the same base, we can equate the exponents: \(\frac{x-3}{4} = 0.5\). From here, solve for the unknown by isolating the variable. Multiply both sides by 4 to remove the fraction and simplify: \(x-3 = 4 \times 0.5\). After that, solve for \(x\) by adding 3 to both sides, giving you the final solution: \(x = 2 + 3\), which simplifies to \(x = 5\). This process demonstrates how tackling the equation step-by-step after expressing it with a common base can lead to a straightforward solution.

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Most popular questions from this chapter

Use the exponential decay model, \(A=A_{0} e^{k t},\) to solve Exercises \(28-31 .\) Round answers to one decimal place. The half-life of thorium- 229 is 7340 years. How long will it take for a sample of this substance to decay to \(20 \%\) of its original amount?

The August 1978 issue of National Geographic described the 1964 find of bones of a newly discovered dinosaur weighing 170 pounds, measuring 9 feet, with a 6 -inch claw on one toe of each hind foot. The age of the dinosaur was estimated using potassium-40 dating of rocks surrounding the bones. a. Potassium- 40 decays exponentially with a half-life of approximately 1.31 billion years. Use the fact that after 1.31 billion years a given amount of potassium-40 will have decayed to half the original amount to show that the decay model for potassium-40 is given by \(A=A_{0} e^{-0.52912 t}\) where \(t\) is in billions of years. b. Analysis of the rocks surrounding the dinosaur bones indicated that \(94.5 \%\) of the original amount of potassium-40 was still present. Let \(A=0.945 A_{0}\) in the model in part (a) and estimate the age of the bones of the dinosaur.

Exercises \(51-56\) present data in the form of tables. For each data set shown by the table, a. Create a scatter plot for the data. b. Use the scatter plot to determine whether an exponential function, a logarithmic function, or a linear function is the best choice for modeling the data. (If applicable, in Exercise \(76,\) you will use your graphing utility to obtain these functions.) Hamachiphobia $$\begin{array}{lcc}\hline & \begin{array}{c} \text { Percentage } \\ \text { Who Won't } \\ \text { Try Sushi } \end{array} & \begin{array}{c} \text { Percentage Who } \\ \text { Don't Approve of } \\ \text { Marriage Equality } \end{array} \\ \hline \text { Millennials } & 42 & 36 \\ \text { Gen X } & 52 & 49 \\ \text { Boomers } & 60 & 59 \\ \text { Silent/Greatest } & 72 & 66 \\ \text { Generation } & & \end{array}$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. After 100 years, a population whose growth rate is \(3 \%\) will have three times as many people as a population whose growth rate is \(1 \%\)

The percentage of adult height attained by a girl who is \(x\) years old can be modeled by $$ f(x)=62+35 \log (x-4) $$ where \(x\) represents the girl's age (from 5 to 15 ) and \(f(x)\) represents the percentage of her adult height. Round answers to the nearest tenth of a percent. Approximately what percentage of her adult height has a girl attained at age ten?
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