Problem 4
Question
If \(x=e^{y},\) then \(y=\) ______.
Step-by-Step Solution
Verified Answer
The solution of the given equation, \(x = e^y\), is \(y = \ln(x)\)
1Step 1: Identify the Problem
We have an exponential equation where 'x' is equal to \(e^y\). We want to solve this equation for 'y'.
2Step 2: Apply Natural Logarithm
To isolate and solve for 'y', we take the natural logarithm (ln) of both sides of the equation. In mathematical terms, ln is the inverse operation of exponentiation. So we write the equation as: \(\ln(x) = \ln(e^y)\)
3Step 3: Use the logarithm identity
There is a logarithmic identity that states \(\ln(a^b)\) is equal to \(b* \ln(a)\). So we can simplify the equation, \(\ln(e^y)\), to \(y * \ln(e)\). \(\ln(e)\) is 1, so the equation simplifies down even further to \(y=\ln(x)\).
Key Concepts
Natural LogarithmsExponential FunctionsLogarithmic Identities
Natural Logarithms
Natural logarithms, often denoted as "ln", are a key tool in solving exponential equations. They serve as the inverse of exponential functions. When you encounter an equation in the form of \(x = e^{y}\), the natural logarithm helps by reversing the exponential action. In essence, applying the natural logarithm to both sides cancels the exponent, thus isolating the variable for which you solve.
The natural logarithm specifically uses base \(e\), which is approximately equal to 2.71828. This particular base is derived from growth processes such as population and radioactive decay, among others. To solve for \(y\) in our example, we take \(ln(x) = ln(e^y)\), allowing us to eventually express \(y\) as \(ln(x)\). This makes it quite a powerful method, especially for real-world applications where such derived constants play a role.
The natural logarithm specifically uses base \(e\), which is approximately equal to 2.71828. This particular base is derived from growth processes such as population and radioactive decay, among others. To solve for \(y\) in our example, we take \(ln(x) = ln(e^y)\), allowing us to eventually express \(y\) as \(ln(x)\). This makes it quite a powerful method, especially for real-world applications where such derived constants play a role.
Exponential Functions
Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In our exercise, we have \(x = e^{y}\), where \(e\) is the base and \(y\) is the exponent. These functions are vital in modeling growth and decay processes because they deftly describe scenarios where quantities double, triple, or halve over consistent intervals of time.
They have distinct properties, such as continuous growth and a constant rate of proportion. In other words, the rate at which the function grows is directly proportional to its current value. This is why exponentials are pivotal in fields like finance and physics. When manipulating these functions, such as trying to solve for an unknown exponent, natural logarithms become essential as they directly counteract the effects of exponentiation.
They have distinct properties, such as continuous growth and a constant rate of proportion. In other words, the rate at which the function grows is directly proportional to its current value. This is why exponentials are pivotal in fields like finance and physics. When manipulating these functions, such as trying to solve for an unknown exponent, natural logarithms become essential as they directly counteract the effects of exponentiation.
Logarithmic Identities
Logarithmic identities are rules that simplify expressions involving logarithms, making them easier to solve or manipulate. One central identity is \(\ln(a^b) = b \cdot \ln(a)\). It states that a logarithm of a power can be turned into a product. This is notably helpful when faced with an expression like \(\ln(e^y)\).
In our context, applying this identity allows us to transform the problem into something more manageable: \(\ln(e^y)\) becomes \(y \cdot \ln(e)\). Since \(\ln(e) = 1\), this expression simplifies neatly to just \(y\). Through such identities, logarithms create a bridge from complex exponential expressions to simpler algebraic equations. Understanding and using these identities are crucial for troubleshooting or solving equations involving logarithms and exponential terms.
In our context, applying this identity allows us to transform the problem into something more manageable: \(\ln(e^y)\) becomes \(y \cdot \ln(e)\). Since \(\ln(e) = 1\), this expression simplifies neatly to just \(y\). Through such identities, logarithms create a bridge from complex exponential expressions to simpler algebraic equations. Understanding and using these identities are crucial for troubleshooting or solving equations involving logarithms and exponential terms.
Other exercises in this chapter
Problem 4
Can you solve \(5^{x}=125\) using a One-to-One Property?
View solution Problem 4
Which property of logarithms can you use to condense the expression \(\ln x-\ln 2 ?\)
View solution Problem 4
The formula \(A=P e^{\pi}\) gives the balance \(A\) of an account earning what type of interest?
View solution Problem 5
Does the model \(y=120 e^{-0.25 x}\) represent exponential growth or exponential decay?
View solution