Problem 18 - Precalculus with Calculus Previews

Question

In Problems \(13-18\), find the exact value of the given logarithm. $$ \ln \left(e^{4} e^{9}\right) $$

Step-by-Step Solution

Answer

The exact value is 13.

Step 1: Apply Logarithm Properties

We start with the expression \( \ln(e^4 e^9) \). You can simplify this using the property of logarithms \( \ln(a \, b) = \ln a + \ln b \): \[ \ln(e^4 e^9) = \ln(e^4) + \ln(e^9) \]

Step 2: Simplify Using the Power Rule

The second step is to apply the power rule of logarithms, which is \( \ln(a^b) = b \ln(a) \). Apply this to both terms: \[ \ln(e^4) = 4 \ln(e) \] and \[ \ln(e^9) = 9 \ln(e) \]

Step 3: Recognize the Natural Logarithm Property

Remember that the natural logarithm \( \ln(e) \) of the base \( e \) is 1. Therefore, you can simplify: \[ 4 \ln(e) = 4 \times 1 = 4 \]\[ 9 \ln(e) = 9 \times 1 = 9 \]

Step 4: Combine the Results

Add the simplified terms together: \[ 4 + 9 = 13 \]

Step 5: Final Step: Present the Exact Value

Therefore, the exact value of the given logarithm \( \ln(e^4 e^9) \) is 13.

Key Concepts

Natural Logarithm

Understanding what a natural logarithm means is crucial when dealing with problems involving logarithms and exponentials. The natural logarithm is represented by the symbol \( \ln \) and is specifically the logarithm taken to the base of \( e \), where \( e \) is an irrational constant approximately equal to \( 2.71828 \). In other terms, when you see \( \ln(x) \), it means you're asking the question, "To what power must \( e \) be raised, to produce \( x \)?"

Natural logarithms are particularly common in processes involving exponential growth or decay, like population growth or radioactive decay. They're also the "natural" choice in the field of calculus, making extensions like integration and differentiation more straightforward than other logarithmic bases. It's important to remember that \( \ln(e) = 1 \) because \( e^1 = e \). This property greatly simplifies computations involving the base \( e \).

• Example: \( \ln(1) = 0 \) because \( e^0 = 1 \)
• Example: \( \ln(e^k) = k \) due to the property that \( \ln(e) = 1 \)

Logarithm Properties

Logarithms possess several useful properties that simplify complex calculations. One vital property is the product rule, which states that the logarithm of a product is the sum of the logarithms of the factors: \( \ln(ab) = \ln a + \ln b \). This rule allows us to break down more complicated logarithmic expressions into simpler sums, making calculations more manageable.

Another important feature is the quotient rule, formulated as \( \ln(a/b) = \ln a - \ln b \). This is especially helpful when dealing with division inside a logarithmic function.

Moreover, the change of base formula is another property, often used to switch the base of a logarithm to one that's more convenient for calculation. It's expressed as \( \log_b(a) = \frac{\ln a}{\ln b} \), which can be useful if you're interested in calculating logarithms with a different base.

• Product Rule: \( \ln(2) + \ln(3) = \ln(6) \)
• Quotient Rule: \( \ln(8) - \ln(2) = \ln(4) \)
• Change of Base: \( \log_2(8) = \frac{\ln 8}{\ln 2} = 3 \)

Power Rule of Logarithms

The power rule is another fundamental property of logarithms that proves exceptionally useful. This rule states that the logarithm of a number raised to an exponent is equal to the exponent times the logarithm of the base number. It is expressed as \( \ln(a^b) = b \ln(a) \).

This is incredibly powerful because it allows you to "pull down" exponents in logarithmic expressions, making them much easier to handle. When the base of the exponent is \( e \), like in the original exercise, this property simplifies calculations significantly, because \( \ln(e) = 1 \).

For example, if you're given \( \ln(e^4) \), using the power rule you can transform this into \( 4 \ln(e) \), and since \( \ln(e) = 1 \), this simplifies further to just \( 4 \).

• Example: \( \ln(x^3) = 3 \ln(x) \)
• Example: \( \ln(e) = 1; \ln(e^5) = 5 \ln(e) = 5 \)

Remember, leveraging the power rule can also help you solve logarithmic equations quickly and with less complexity.