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Solve the exponential equation. Round to three decimal places, when needed. $$.5 e^{x}=60$$

Short Answer

Expert verified
The solution to the equation is \(x \approx 4.787\).

Step by step solution

01

Isolate the exponential portion of the equation

First, we want to isolate the term with the exponent on one side of the equation. We do this by simply dividing both sides of the equation by 0.5. That leaves us with the equation \(e^{x} = 60 / 0.5\), which simplifies to \(e^{x} = 120\).
02

Apply natural logarithm to both sides

Next, we will apply the natural logarithm (ln) to both sides of the equation to get rid of the base 'e'. This will result in \(x = \ln(120)\). We apply log because of the property of logarithms that \(\ln(e^{x}) = x\).
03

Calculate the value

Now, all we need to do is to calculate the natural logarithm of 120. Using a calculator or a logarithm table, we find that \(\ln(120) \approx 4.787\). However, we need the result to be rounded to three decimal places as per the instruction in the exercise. Therefore, \(x \approx 4.787\).

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Key Concepts

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

Understanding Natural Logarithms
Natural logarithms are an essential tool in solving exponential equations, particularly those involving the base 'e'. The natural logarithm, commonly denoted as \( \ln \), is the inverse function of the exponential function with base \( e \). This means applying a natural logarithm to \( e^{x} \) will give us \( x \).

The term "natural" arises because the base \( e \) is approximately 2.71828, a unique number related to growth processes seen in nature such as compounded interest or certain types of exponential decay.

When you have an exponential equation like \( e^{x} = 120 \), using the natural logarithm on both sides simplifies the equation to \( x = \ln(120) \). This makes \( \ln \) a powerful tool because it simplifies solving for \( x \) without requiring complicated algebra.

Remember:
  • \( \ln(e^{x}) = x \)
  • It changes exponential expressions to linear ones.
It's especially handy when the exponent is the variable we're solving for.
Isolation of Variables in Equations
In mathematics, isolating the variable is the crucial initial step when solving equations. It involves rearranging the equation so that the variable you're interested in (like \( x \) in our example equation) appears by itself on one side of the equation.

For the given exercise, the equation was initially \( 0.5 e^{x}=60 \). To isolate the exponential term \( e^{x} \), you divide both sides by 0.5, resulting in \( e^{x} = 120 \). This step separates the variable term from any coefficients or other terms, hence simplifying the equation.

Key tips:
  • Isolate the term containing the variable first.
  • Perform operations (like division) on both sides to maintain equality.
Ensuring that the variable is isolated allows you to then apply other mathematical techniques, such as logarithms, to further simplify and solve the equation.
Properties of Logarithms
Logarithms have many properties that make them indispensable in solving exponential equations. Some of the most important properties that we use in mathematical calculations are particularly helpful in transforming the variables of the equation.

Properties of logarithms include:
  • The property \( \ln(e^{x}) = x \), which we used in solving the equation by applying \( \ln \) to both sides to cancel out \( e \) and get \( x \).
  • Logarithm power rule: \( \ln(a^{b}) = b \cdot \ln(a) \).
  • Addition rule: \( \ln(a) + \ln(b) = \ln(ab) \).
  • Subtraction rule: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \).
By understanding these properties, you can tackle a wide range of problems involving logarithms and exponential expressions. They help to not only solve but also simplify complex expressions, making them easier to manage.

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