Problem 13
Question
Find the solution of the exponential equation, correct to four decimal places. $$ 8^{0.4 x}=5 $$
Step-by-Step Solution
Verified Answer
The value of x is approximately 1.9354.
1Step 1: Take the natural logarithm on both sides
Rewrite the equation as \(8^{0.4x} = 5\). To solve for \(x\), take the natural logarithm on both sides: \(\ln(8^{0.4x}) = \ln(5)\). This helps to bring down the exponent using the properties of logarithms.
2Step 2: Apply the power rule for logarithms
Using the power rule for logarithms, which states that \(\ln(a^b) = b \cdot \ln(a)\), simplify the left side: \(0.4x \cdot \ln(8) = \ln(5)\).
3Step 3: Solve for x
Isolate \(x\) by dividing both sides of the equation by \(0.4 \cdot \ln(8)\). The equation becomes \(x = \frac{\ln(5)}{0.4 \cdot \ln(8)}\). Calculate both ln(5) and ln(8): \(\ln(5) \approx 1.6094\) and \(\ln(8) \approx 2.0794\). Substitute these values into the equation.
4Step 4: Perform the calculation
Substitute the logarithm values into the equation \(x = \frac{1.6094}{0.4 \cdot 2.0794}\). Simplify this calculation: \(x \approx \frac{1.6094}{0.83176}\). Divide to find \(x\) approximately = 1.9354.
Key Concepts
Natural LogarithmsPower Rule for LogarithmsSolving for x
Natural Logarithms
When solving exponential equations, one of the essential tools used is the natural logarithm, often represented as \( \ln \). It is the logarithm to the base \( e \), where \( e \approx 2.71828 \). This function is particularly useful in solving equations like the one in our exercise because natural logarithms allow us to transform exponential terms into more manageable forms.
The natural logarithm helps in simplifying problems that involve growth or decay processes, compounded interest, and many naturally occurring phenomena. The key property that we use when working with natural logarithms is that they provide a means to 'bring down' the exponent in an expression of the form \( a^b \).
The natural logarithm helps in simplifying problems that involve growth or decay processes, compounded interest, and many naturally occurring phenomena. The key property that we use when working with natural logarithms is that they provide a means to 'bring down' the exponent in an expression of the form \( a^b \).
- It turns multiplication into addition: \( \ln(a \cdot b) = \ln(a) + \ln(b) \).
- It transforms division into subtraction: \( \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b) \).
- It simplifies powers: \( \ln(a^b) = b \cdot \ln(a) \).
Power Rule for Logarithms
The power rule for logarithms is a pillar of logarithmic properties, allowing us to manipulate expressions with exponents. This rule states that \( \ln(a^b) = b \cdot \ln(a) \), meaning we can move the exponent "in front" of the logarithm, making it a multiplier.
In the context of solving exponential equations, this rule is invaluable. It allows us to take expressions where the variable we want to solve for is in the exponent and bring it down into a linear form, greatly simplifying our task.
Let's apply this to our exercise:
In the context of solving exponential equations, this rule is invaluable. It allows us to take expressions where the variable we want to solve for is in the exponent and bring it down into a linear form, greatly simplifying our task.
Let's apply this to our exercise:
- Start with the equation \( \ln(8^{0.4x}) = \ln(5) \).
- Using the power rule, rewrite the left side as \( 0.4x \cdot \ln(8) \).
- This results in the equation \( 0.4x \cdot \ln(8) = \ln(5) \).
Solving for x
Once the equation is simplified using logarithmic properties, the final step involves isolating \( x \). From our exercise, after applying the power rule, we have the equation:
- \( 0.4x \cdot \ln(8) = \ln(5) \).
- First, divide both sides by \( 0.4 \cdot \ln(8) \) to isolate \( x \): \( x = \frac{\ln(5)}{0.4 \cdot \ln(8)} \).
- Calculate the natural logarithms: \( \ln(5) \approx 1.6094 \) and \( \ln(8) \approx 2.0794 \).
- Substitute these values into the equation: \( x = \frac{1.6094}{0.4 \cdot 2.0794} \).
- Simplifying: \( x \approx 1.9354 \).
Other exercises in this chapter
Problem 12
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