Problem 64
Question
In Exercises, solve for \(x\) or \(t\). $$ 2 e^{-x+1}-5=9 $$
Step-by-Step Solution
Verified Answer
The solution for \(x\) is \(x = 1 - \ln(7)\).
1Step 1: Isolate the exponential term
We start by isolating the term with the \(e\) value on one side of the equation. Begin with the equation: \(2 e^{-x+1} - 5 = 9\). Add 5 to both sides of the equation to get: \(2 e^{-x+1} = 9 + 5\), which simplifies to: \(2 e^{-x+1} = 14\).
2Step 2: Remove coefficient of exponential term
Dividing both sides of the equation by 2 gives: \(e^{-x+1} = 14/2\), which simplifies to: \(e^{-x+1} = 7\).
3Step 3: Use natural logarithm
Next, apply the natural logarithm to both sides of the equation (note that the natural log, \(ln\), is the inverse operation to \(e\)). This is written as: \(ln(e^{-x+1}) = ln(7)\). The property of logarithms \(\ln(a^b) = b \ln(a)\) can now be applied to the LHS to obtain: \(-x+1 = \ln(7)\).
4Step 4: Solve for \(x\)
Finally, solve for \(x\) by isolating \(x\) on one side to obtain: \(x = 1 - \ln(7)\).
Key Concepts
Natural LogarithmIsolating VariablesSolving Equations
Natural Logarithm
The natural logarithm, often denoted as \(\ln\), is a logarithm with the base \(e\), where \(e\) is approximately equal to 2.71828. It is a widely used concept in mathematics, especially in calculus and complex equations involving exponential functions.
The natural logarithm serves as the inverse function of the exponential function with base \(e\). This means that if you have an equation like \(e^y = x\), applying the natural logarithm to both sides gives \(y = \ln(x)\). This property is extremely useful in solving exponential equations, allowing us to switch from an exponential to a linear format.
When dealing with natural logarithms, there are important properties to remember:
The natural logarithm serves as the inverse function of the exponential function with base \(e\). This means that if you have an equation like \(e^y = x\), applying the natural logarithm to both sides gives \(y = \ln(x)\). This property is extremely useful in solving exponential equations, allowing us to switch from an exponential to a linear format.
When dealing with natural logarithms, there are important properties to remember:
- \(\ln(1) = 0\) because \(e^0 = 1\).
- \(\ln(e) = 1\) because \(e^1 = e\).
- \(\ln(a^b) = b \ln(a)\), which is particularly useful when we need to simplify expressions involving exponents.
Isolating Variables
Isolating a variable means rearranging an equation so that a particular variable is on one side of the equation by itself. This process is fundamental in solving algebraic equations.
In the given equation \(2 e^{-x+1} - 5 = 9\), our goal is to isolate the exponential term, \(e^{-x+1}\), which involves a series of strategic algebraic manipulations:
In the given equation \(2 e^{-x+1} - 5 = 9\), our goal is to isolate the exponential term, \(e^{-x+1}\), which involves a series of strategic algebraic manipulations:
- First, eliminate any constant that is being added or subtracted from the term we want to isolate. Here, adding 5 to both sides of the equation gives \(2 e^{-x+1} = 14\).
- Next, if there's a coefficient multiplying the term, divide it on both sides. For our equation, dividing by 2 isolates \(e^{-x+1}\), providing us \(e^{-x+1} = 7\).
Solving Equations
Solving equations involves finding the value of the variable that makes the equation true. This requires understanding the operations involved and applying appropriate mathematical principles.
In our problem, solving for \(x\) required the following steps:
1. **Isolate the Exponential:** Start with the equation \(2 e^{-x+1} - 5 = 9\). This was done by isolating the exponential part, simplifying it to \(e^{-x+1} = 7\) through basic algebraic steps as outlined in the previous section.
2. **Apply the Natural Logarithm:** Once the exponential term is isolated, apply the natural logarithm to both sides: \(\ln(e^{-x+1}) = \ln(7)\). This uses the principle that the natural logarithm is the inverse of the exponential function, thus simplifying the equation to \(-x + 1 = \ln(7)\).
3. **Rearrange to Solve for \(x\):** Finally, solve for \(x\) by isolating it: \(x = 1 - \ln(7)\). This step is straightforward since we are now dealing with a simple linear equation.
These steps illustrate the entire process from recognizing the form of the equation to applying algebraic and logarithmic properties effectively, culminating in the correct solution.
In our problem, solving for \(x\) required the following steps:
1. **Isolate the Exponential:** Start with the equation \(2 e^{-x+1} - 5 = 9\). This was done by isolating the exponential part, simplifying it to \(e^{-x+1} = 7\) through basic algebraic steps as outlined in the previous section.
2. **Apply the Natural Logarithm:** Once the exponential term is isolated, apply the natural logarithm to both sides: \(\ln(e^{-x+1}) = \ln(7)\). This uses the principle that the natural logarithm is the inverse of the exponential function, thus simplifying the equation to \(-x + 1 = \ln(7)\).
3. **Rearrange to Solve for \(x\):** Finally, solve for \(x\) by isolating it: \(x = 1 - \ln(7)\). This step is straightforward since we are now dealing with a simple linear equation.
These steps illustrate the entire process from recognizing the form of the equation to applying algebraic and logarithmic properties effectively, culminating in the correct solution.
Other exercises in this chapter
Problem 63
In Exercises, solve for \(x\) or \(t\). $$ 4 e^{2 x-1}-1=5 $$
View solution Problem 63
In Exercises, find the second derivative of the function. $$ f(x)=5^{x} $$
View solution Problem 64
In Exercises, find the second derivative of the function. $$ f(x)=\log _{10} x $$
View solution Problem 65
In Exercises, solve for \(x\) or \(t\). $$ \frac{10}{1+4 e^{-0.01 x}}=2.5 $$
View solution