Evaluating Natural Logarithm & Exponential Expressions

Let's dive into evaluating some expressions involving natural logarithms and exponential functions without reaching for a calculator. We'll break down each expression step-by-step, leveraging the fundamental properties of logarithms and exponentials to arrive at the solutions.

(a) ${ \ln \left(\frac{1}{e^5}\right) }$

When dealing with the natural logarithm of a fraction where the denominator is an exponential term, it's essential to recall the properties of logarithms and exponents. The expression ${ \ln \left(\frac{1}{e^5}\right) }$ can be simplified using the logarithm property that states ${ \ln \left(\frac{1}{x}\right) = -\ln(x) }$. Applying this property, we rewrite the expression as:

${ \ln \left(\frac{1}{e^5}\right) = -\ln(e^5) }$

Now, we need to address ${ \ln(e^5) }$. Remember that the natural logarithm, denoted as ${ \ln }$, is the logarithm to the base e. Therefore, ${ \ln(e^x) = x }$ due to the inverse relationship between the natural logarithm and the exponential function. Applying this inverse relationship, we get:

${ -\ln(e^5) = -5 }$

Thus, the simplified value of the original expression is -5. To summarize, we first used the reciprocal property of logarithms to transform the fraction into a more manageable form. Then, we applied the inverse relationship between natural logarithms and exponential functions to directly find the exponent. This process showcases the importance of recognizing and applying logarithmic and exponential properties for efficient simplification. Understanding these properties allows us to bypass complex calculations and arrive at the solution directly, which is particularly useful in scenarios where calculators are not available.

(b) ${ e^{\ln (6)} }$

In this part, we are asked to evaluate the expression ${ e^{\ln (6)} }$. This expression involves an exponential function with a natural logarithm in the exponent. The key to simplifying this expression lies in understanding the inverse relationship between the exponential function (base e) and the natural logarithm. By definition, the natural logarithm ${ \ln(x) }$ is the power to which e must be raised to obtain x. Therefore, ${ e^{\ln (x)} }$ simplifies directly to x.

Applying this inverse relationship to our expression, we have:

${ e^{\ln (6)} = 6 }$

The simplification is quite straightforward: the exponential function ${ e }$ and the natural logarithm ${ \ln }$ effectively cancel each other out, leaving the argument of the logarithm, which is 6, as the result. This principle is fundamental in simplifying expressions that combine exponential and logarithmic functions. Understanding this inverse relationship is crucial because it allows for quick and efficient evaluation without needing to perform complex calculations. It’s a direct application of the definition of logarithms and exponentials, making it a powerful tool in mathematical evaluations.

The simplicity of this evaluation highlights the elegance of mathematical relationships when correctly applied. Recognizing these relationships not only saves time but also reinforces a deeper understanding of the underlying mathematical principles.

(c) ${ e^{\ln (\sqrt{10})} }$

The expression we need to evaluate here is ${ e^{\ln (\sqrt{10})} }$. Similar to the previous example, this involves an exponential function with a natural logarithm in the exponent. The critical concept to remember is the inverse relationship between the exponential function with base e and the natural logarithm function. Specifically, ${ e^{\ln(x)} = x }$.

In our case, x is ${ \sqrt{10} }$. Applying the inverse relationship directly, we have:

${ e^{\ln (\sqrt{10})} = \sqrt{10} }$

Therefore, the simplified form of the expression ${ e^{\ln (\sqrt{10})} }$ is simply ${ \sqrt{10} }$. This result underscores the utility of recognizing inverse functions in mathematical expressions. When you see an exponential function ${ e }$ raised to the power of a natural logarithm ${ \ln }$, they effectively cancel each other out, leaving behind the argument of the logarithm. This is a direct consequence of how logarithms and exponentials are defined and related to each other.

This type of problem reinforces the idea that understanding fundamental mathematical relationships can greatly simplify complex-looking expressions. By recognizing and applying the inverse relationship, we avoid the need for approximation or calculator use, providing an exact solution directly.

(d) ${ \ln \left(e^{-3}\right) }$

Evaluate the expression ${ \ln \left(e^{-3}\right) }$. This expression involves the natural logarithm of an exponential term. To simplify this, we rely on the fundamental property of logarithms, which states that ${ \ln(e^x) = x }$, where ${ \ln }$ is the natural logarithm (i.e., logarithm to the base e). This property arises directly from the inverse relationship between exponential and logarithmic functions.

Applying this property to the expression, we have:

${ \ln \left(e^{-3}\right) = -3 }$

Thus, the natural logarithm of ${ e^{-3} }$ simplifies directly to -3. The natural logarithm essentially "undoes" the exponential function, leaving the exponent as the result. This is a key concept in understanding how logarithms and exponentials interact.

This type of simplification is crucial in many areas of mathematics and science, particularly when dealing with exponential decay or growth models. Recognizing this property allows for quick and efficient simplification, making it easier to solve equations and understand the behavior of various functions. It emphasizes the importance of understanding the inverse relationships between different mathematical operations.

(e) ${ \ln \left(\sqrt{e^5}\right) }$

We aim to evaluate the expression ${ \ln \left(\sqrt{e^5}\right) }$. This involves the natural logarithm of a square root of an exponential term. To simplify this expression, we will use properties of both exponents and logarithms. First, recall that a square root can be represented as an exponent of ${ \frac{1}{2} }$. Therefore, we can rewrite the expression as:

${ \ln \left(\sqrt{e^5}\right) = \ln \left((e^5)^{\frac{1}{2}}\right) }$

Next, we use the power of a power rule, which states that ${ (a^m)^n = a^{mn} }$. Applying this rule, we get:

${ \ln \left((e^5)^{\frac{1}{2}}\right) = \ln \left(e^{\frac{5}{2}}\right) }$

Now, we apply the property of logarithms that ${ \ln(e^x) = x }$. In this case, ${ x = \frac{5}{2} }$, so we have:

${ \ln \left(e^{\frac{5}{2}}\right) = \frac{5}{2} }$

Thus, the simplified value of the expression ${ \ln \left(\sqrt{e^5}\right) }$ is ${ \frac{5}{2} }$. This process involves rewriting the square root as a fractional exponent, applying the power of a power rule to simplify the exponent, and then using the inverse relationship between the natural logarithm and the exponential function. Each step relies on understanding and applying fundamental properties, illustrating how complex expressions can be simplified through careful application of mathematical rules.

In summary, evaluating expressions involving logarithms and exponentials often requires a solid understanding of their properties and relationships. Recognizing these properties allows for efficient simplification and accurate evaluation without the need for computational aids. Practicing these types of problems reinforces these fundamental concepts, making them easier to apply in more complex scenarios.

For further reading on logarithms and exponential functions, you might find helpful resources on Khan Academy. This website offers comprehensive lessons and practice exercises to deepen your understanding of these mathematical concepts.