Problem 16
Question
In Exercises \(13-16,\) solve for \(t\). $$ e^{\left(x^{2}\right)} e^{(2 x+1)}=e^{t} $$
Step-by-Step Solution
Verified Answer
\(t = x^2 + 2x + 1\)
1Step 1: Identify the Equation Components
The given equation is \(e^{(x^2)} e^{(2x+1)} = e^{t}\). This can be recognized as an equation with exponential expressions on both sides.
2Step 2: Combine Exponentials on the Left Side
Using the properties of exponents, specifically that \(e^a \times e^b = e^{(a+b)}\), we can combine the left side as follows: \(e^{(x^2)} e^{(2x+1)} = e^{(x^2 + 2x + 1)}\).
3Step 3: Equate the Exponents
Since the bases are the same (all are \(e\)), we can equate the exponents. Therefore, \(x^2 + 2x + 1 = t\). This gives us the equation in the form needed to solve for \(t\).
4Step 4: Solve for \(t\)
The equation \(t = x^2 + 2x + 1\) is already solved for \(t\). This represents the relationship between \(t\) and \(x\).
Key Concepts
Properties of ExponentsSolving EquationsEquivalent Exponents
Properties of Exponents
Understanding the properties of exponents is crucial when dealing with exponential equations. One key property to remember is that when you multiply exponential expressions with the same base, you can add the exponents. For instance, if you have expressions like \(e^a \times e^b\), you can simplify them to \(e^{(a+b)}\).
\[e^a \times e^b = e^{a+b}\]
This property allows you to handle complicated multiplications of exponents more easily by turning them into addition, which is usually simpler to compute. This can be very helpful in various mathematical problems involving exponential growth or decay.
Other important properties include raising a power to a power, such as \((e^a)^b = e^{ab}\), and applying exponential rules to fractions and negative exponents. This toolbox of properties provides you with flexible strategies when solving equations, especially in mathematics and science problems.
\[e^a \times e^b = e^{a+b}\]
This property allows you to handle complicated multiplications of exponents more easily by turning them into addition, which is usually simpler to compute. This can be very helpful in various mathematical problems involving exponential growth or decay.
Other important properties include raising a power to a power, such as \((e^a)^b = e^{ab}\), and applying exponential rules to fractions and negative exponents. This toolbox of properties provides you with flexible strategies when solving equations, especially in mathematics and science problems.
Solving Equations
Solving equations involving exponents often requires rewriting the equation in simpler forms, using properties mentioned above, to find the variable. In our example with exponential terms \(e^{(x^2)}e^{(2x+1)} = e^{t}\), the first step is to combine the exponents on the left side using the addition property.
This simplifies our equation to \(e^{(x^2 + 2x + 1)} = e^t\).
The next crucial step is recognizing that if you have two exponential expressions with the same base (like \(e\) in this case), this allows you to directly equate their exponents. Therefore, \(x^2 + 2x + 1 = t\) simplifies your task of finding the unknown in the equation. Breaking down equations like this to their simplest forms allows for a much clearer path to the solution, minimizing potential errors in complex expressions.
This simplifies our equation to \(e^{(x^2 + 2x + 1)} = e^t\).
The next crucial step is recognizing that if you have two exponential expressions with the same base (like \(e\) in this case), this allows you to directly equate their exponents. Therefore, \(x^2 + 2x + 1 = t\) simplifies your task of finding the unknown in the equation. Breaking down equations like this to their simplest forms allows for a much clearer path to the solution, minimizing potential errors in complex expressions.
Equivalent Exponents
The concept of equivalent exponents involves recognizing when two exponential expressions can be compared or equated. As shown in the example, after using properties of exponents to simplify the equation, you end up with \(e^{(x^2 + 2x + 1)} = e^t\). Since the bases (in this case, \(e\)) are identical, the exponents themselves must be equivalent.
This means that you can set \(x^2 + 2x + 1 = t\) directly, converting the problem to solving a polynomial expression in terms of \(x\).
This is because exponential functions are one-to-one, meaning they produce a unique output for each unique input. So, if two outputs of an exponential function with the same base are equal, their inputs (the exponents) must be equal as well. Understanding this can simplify dealing with complex equations and make finding solutions much more manageable.
This means that you can set \(x^2 + 2x + 1 = t\) directly, converting the problem to solving a polynomial expression in terms of \(x\).
This is because exponential functions are one-to-one, meaning they produce a unique output for each unique input. So, if two outputs of an exponential function with the same base are equal, their inputs (the exponents) must be equal as well. Understanding this can simplify dealing with complex equations and make finding solutions much more manageable.
Other exercises in this chapter
Problem 16
In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=t^{2} \tanh \frac{1}{t} $$
View solution Problem 16
Given that \(\alpha=\sec ^{-1}(-\sqrt{13} / 2),\) find \(\sin \alpha, \cos \alpha, \tan \alpha, \csc \alpha\) and \(\cot \alpha .\)
View solution Problem 16
(Continuation of Exercise \(15 .\) ) In Benjamin Franklin's estimate that the original 1000 pounds would grow to \(131,000\) in 100 years, he was using an annua
View solution Problem 16
(Continution of Exercise \(15 . )\) Show that the value of $$ \lim _{x \rightarrow \infty} \frac{\ln (x+a)}{\ln x} $$ is the same no matter what value you assig
View solution