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Chapter 5: Problem 46

Solve the equations. Express the answers in terms of natural logarithms. $$5^{3 x-1}=27$$

Short Answer

Expert verified
\( x = \frac{1}{3} \left( \frac{3\ln(3)}{\ln(5)} + 1 \right) \)

Step by step solution

01

Take the Natural Logarithm of Both Sides

To solve the equation \( 5^{3x-1} = 27 \), we first take the natural logarithm (ln) of both sides to help bring down the exponent. This gives us \( \ln(5^{3x-1}) = \ln(27) \).
02

Use the Power Rule of Logarithms

Apply the power rule of logarithms, which states that \( \ln(a^b) = b \cdot \ln(a) \). Thus, \( (3x-1) \cdot \ln(5) = \ln(27) \).
03

Solve for the Variable x

To isolate \( x \), first divide both sides by \( \ln(5) \): \[ 3x - 1 = \frac{\ln(27)}{\ln(5)} \]Next, add 1 to both sides:\[ 3x = \frac{\ln(27)}{\ln(5)} + 1 \]Finally, divide everything by 3:\[ x = \frac{1}{3} \left( \frac{\ln(27)}{\ln(5)} + 1 \right) \]
04

Express the Final Answer in Simplified Form

Since 27 can be rewritten as \( 3^3 \), we can simplify \( \ln(27) \) to \( 3\ln(3) \). Therefore, the expression becomes:\[ x = \frac{1}{3} \left( \frac{3\ln(3)}{\ln(5)} + 1 \right) \]This is the solution expressed in terms of natural logarithms.

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Key Concepts

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

Understanding Exponentiation
Exponentiation is a mathematical operation that represents repeated multiplication. It involves a base number and an exponent. For example, in the expression \( 5^{3x-1} \), the number 5 is the base, and \( 3x-1 \) is the exponent. This means that the number 5 is multiplied by itself \( 3x-1 \) times.

Key concepts of exponentiation include:
  • Base: The number being multiplied.
  • Exponent: The power to which the base is raised.
  • Exponential Form: The expression \( a^b \) indicates a multiplied b times.
Understanding exponentiation is crucial because it simplifies expressions that involve large numbers. It also forms the basis for understanding logarithms, which are inversely related to exponentiation.
Solving Logarithmic Equations
Logarithmic equations involve logarithms, which are the inverse functions of exponentiation. They help us solve equations where the variable is in the exponent, like \( 5^{3x-1} = 27 \). By taking the natural logarithm (\( \ln \)) of both sides of the equation, we transform the expression, allowing us to simplify and solve it.

Logarithm properties such as:
  • Power Rule: \( \ln(a^b) = b \cdot \ln(a) \) allows us to bring the exponent down.
  • Product, Quotient, and Change of Base Rules: Help simplify expressions further.
In this case, by applying the power rule, the equation becomes a linear one, which is straightforward to solve using basic algebra. This methodical approach is essential for finding solutions in terms of natural logarithms.
Finding Algebraic Solutions
Once we've simplified the equation with logarithms, we are left with a more familiar algebraic equation, such as \( 3x - 1 = \frac{\ln(27)}{\ln(5)} \). Solving algebraic equations involves isolating the variable on one side.

The steps include:
  • Divide both sides by a term to simplify the expression for the variable.
  • Add or subtract constants to isolate the variable.
  • Finally, divide or multiply as needed to solve for the variable.
For example, in the equation \( 3x = \frac{\ln(27)}{\ln(5)} + 1 \), we add 1 to both sides and then divide by 3 to solve for \( x \). This systematic approach helps us express complex equations in terms of natural logarithms with clarity and precision, providing a clear pathway to the solution.

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Most popular questions from this chapter

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$\ln \frac{3 x-2}{4 x+1}>\ln 4$$

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$\frac{2}{3}\left(1-e^{-x}\right) \leq-3$$

Let \(f(x)=\ln (x+\sqrt{x^{2}+1}) .\) Find \(f^{-1}(x).\)

(a) Specify the domain of the function \(y=\ln x+\ln (x+2).\) (b) Solve the inequality \(\ln x+\ln (x+2) \leq \ln 35.\)

(a) Suppose that a certain country violates the ban against above-ground nuclear testing and, as a result, an island is contaminated with debris containing the radioactive substance iodine-131. A team of scientists from the United Nations wants to visit the island to look for clues in determining which country was involved. However, the level of radioactivity from the iodine- 131 is estimated to be 30,000 times the safe level. Approximately how long must the team wait before it is safe to visit the island? The half-life of iodine- 131 is 8 days. (b) Rework part (a), assuming instead that the radioactive substance is strontium-90 rather than iodine-131. The half-life of strontium- 90 is 28 years. Assume, as before, that the initial level of radioactivity is 30,000 times the safe level. (This exercise underscores the difference between a half-life of 8 days and one of 28 years.)
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