Chapter 6: Problem 88
If \(5^{-x}=3,\) what does \(5^{3 x}\) equal?
Short Answer
Expert verified
\(5^{3x} = \frac{1}{27}\)
Step by step solution
01
- Rewrite the Equation
The given equation is \(5^{-x} = 3\). First, take the logarithm of both sides to solve for \(-x\).
02
- Take the Logarithm
Apply the natural logarithm (ln) to both sides: \(\ln(5^{-x}) = \ln(3)\).
03
- Simplify Using Logarithmic Properties
Using the property of logarithms that \(\ln(a^b) = b \cdot \ln(a)\), rewrite the left side: \(-x \cdot \ln(5) = \ln(3)\).
04
- Solve for \(x\)
Divide both sides by \(-\ln(5)\) to isolate \(x\): \(x = -\frac{\ln(3)}{\ln(5)}\).
05
- Find \(3x\)
Multiply \(x\) by 3: \(3x = 3 \left( -\frac{\ln(3)}{\ln(5)} \right) = -3 \frac{\ln(3)}{\ln(5)}\).
06
- Find \(5^{3x}\)
Raise 5 to the power of \(3x\): \(5^{3x} = 5^{-\frac{3 \ln(3)}{\ln(5)}}\). Using the property \(5^{\ln_b(a)} = a\) where \(b\) is the base and \(a\) is the antilogarithm, we rewrite it as \(\left(3^{\ln_5(5)}\right)^3 = 3^{-3} = \frac{1}{3^3} = \frac{1}{27}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
logarithmic properties
Logarithms are incredibly useful in solving equations involving exponents. A logarithm tells you the power to which a number must be raised to obtain another number. For instance, \( \text{log}_b(a) = c \) means that \( b^c = a \).
There are several key properties of logarithms that often come in handy:
In the given problem, we apply the power property to simplify logarithms: \( \text{ln}(5^{-x}) \) becomes \( -x \text{ln}(5) \). This step is key to isolating the variable. Always remember these properties to transform and solve logarithmic equations more easily.
There are several key properties of logarithms that often come in handy:
- \textbf{Product Property}: \( \text{log}_b(xy) = \text{log}_b(x) + \text{log}_b(y) \).
- \textbf{Quotient Property}: \( \text{log}_b \frac{x}{y} = \text{log}_b(x) - \text{log}_b(y) \).
- \textbf{Power Property}: \( \text{log}_b(x^y) = y \times \text{log}_b(x) \).
In the given problem, we apply the power property to simplify logarithms: \( \text{ln}(5^{-x}) \) becomes \( -x \text{ln}(5) \). This step is key to isolating the variable. Always remember these properties to transform and solve logarithmic equations more easily.
exponent rules
Understanding exponent rules is essential to manipulate expressions involving powers. Exponents tell us how many times to use a number in a multiplication.
Here are some basic rules:
In our problem, using negative exponents and power rules helps. For example, \( 5^{-x} = 3 \) translates to \( x = -\frac{\text{ln}(3)}{\text{ln}(5)} \) when taking the natural logarithm. Later, we find \( 5^{3x} = \frac{1}{27} \) using these rules effectively.
Here are some basic rules:
- \textbf{Product of Powers}: \( a^m \times a^n = a^{m+n} \) – When you multiply two exponents with the same base, you add their powers.
- \textbf{Quotient of Powers}: \( \frac{a^m}{a^n} = a^{m-n} \) – When you divide two exponents with the same base, you subtract the denominator’s exponent from the numerator’s exponent.
- \textbf{Power of a Power}: \( (a^m)^n = a^{m \times n} \) – When you raise an exponent to another power, you multiply the exponents.
- \textbf{Negative Exponent}: \( a^{-m} = \frac{1}{a^m} \) – A negative exponent indicates a reciprocal.
In our problem, using negative exponents and power rules helps. For example, \( 5^{-x} = 3 \) translates to \( x = -\frac{\text{ln}(3)}{\text{ln}(5)} \) when taking the natural logarithm. Later, we find \( 5^{3x} = \frac{1}{27} \) using these rules effectively.
solving equations
Solving equations involves finding the value of variables that make an equation true. For logarithmic and exponential equations, specific strategies are used:
The approach in the exercise used these steps. Applying logarithms linearizes the equation, solving for \( x \) and then using exponent rules to find \( 5^{3x} \). Practicing these methods will improve your ability to tackle similar problems.
- \textbf{Isolate the Logarithm or Exponential Term}: Start by getting the logarithm or the base with its power alone on one side of the equation.
- \textbf{Apply Logarithms to Both Sides}: This step helps to linearize exponential terms. For example, applying \( \text{ln} \) to \( 5^{-x} = 3 \) simplifies to \( -x \text{ln}(5) = \text{ln}(3) \).
- \textbf{Use Logarithmic Properties or Exponent Rules}: Employ the relevant properties or rules to further simplify and solve for the variable.
- \textbf{Back-substitute if Necessary}: In multistep problems, solve for intermediate variables (like \( x = -\frac{\text{ln}(3)}{\text{ln}(5)} \)) and then substitute back to find the final answer (\( 5^{3x} \)).
The approach in the exercise used these steps. Applying logarithms linearizes the equation, solving for \( x \) and then using exponent rules to find \( 5^{3x} \). Practicing these methods will improve your ability to tackle similar problems.