Problem 15 - Thomas Calculus

Question

In Exercises \(13-16,\) solve for \(t\). $$ e^{\sqrt{t}}=x^{2} $$

Step-by-Step Solution

Answer

The solution for \( t \) is \( t = 4(\ln(x))^2 \).

Step 1: Rewrite the Equation Using Logarithms

Start by taking the natural logarithm of both sides of the equation to isolate the exponent. This leads to: \[ \ln(e^{\sqrt{t}}) = \ln(x^2)\] Using the property of logarithms \( \ln(a^b) = b \ln a \), this simplifies to:\[ \sqrt{t} = \ln(x^2)\] Since \( \ln(e) = 1 \).

Step 2: Simplify the Right Side Using Logarithmic Identities

Continue by simplifying the right side using the identity \( \ln(a^b) = b \ln a \). This gives:\[ \sqrt{t} = 2\ln(x)\]

Step 3: Isolate the Square Root

To isolate \( \sqrt{t} \), note that it equals \( 2\ln(x) \) from step 2.

Step 4: Solve for \( t \)

To solve for \( t \), square both sides of the equation:\[ (\sqrt{t})^2 = (2\ln(x))^2\] Simplifying gives:\[ t = 4(\ln(x))^2\] This is the solution for \( t \) in terms of \( x \).

Key Concepts

Natural Logarithms

Natural logarithms, typically denoted as \( \ln \), are logarithms with the base \( e \), where \( e \approx 2.71828 \). This mathematical constant is crucial in calculus and many mathematical calculations. The natural logarithm helps in solving exponential equations by allowing us to bring down exponents. For instance, in the equation \( e^{\sqrt{t}} = x^2 \), taking the natural log of both sides results in \( \ln(e^{\sqrt{t}}) = \ln(x^2) \). Here, \( \ln \) helps transform the equation, paving the way to solve for \( t \).

A key property of natural logarithms is that \( \ln(e^x) = x \). This identity results from the fact that the natural logarithm is the inverse function of exponentiation with base \( e \). Thus, taking the natural log and applying this property simplifies expressions, especially when tackling exponential equations.

Logarithmic Identities

Logarithmic identities are powerful tools in simplifying and solving logarithmic equations. One essential identity is \( \ln(a^b) = b\ln(a) \). This identity allows us to bring down exponents to simplify equations further. Applying this step in our equation, we transition from \( \ln(e^{\sqrt{t}}) = \ln(x^2) \) to \( \sqrt{t} = 2\ln(x) \). Here, the exponent 2 is brought out of the logarithm on the right side.

Other useful logarithmic identities include:

• \( \ln(1) = 0 \) since any number raised to zero is 1.
• \( \ln(xy) = \ln(x) + \ln(y) \) which makes multiplying inside the logarithm equivalent to adding outside.
• \( \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \) meaning division inside the logarithm is subtraction outside.

The proper application of these identities will simplify complex expressions and make solving logarithmic equations much more manageable.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying expressions and equations to solve for unknowns. It involves operations such as addition, subtraction, multiplication, division, and taking roots and powers, which are necessary in resolving exponential and logarithmic equations.

In the context of our exercise, once we obtained \( \sqrt{t} = 2\ln(x) \), the next step was to isolate \( t \). By squaring both sides, we have \( t = (2\ln(x))^2 \), which further simplifies to \( t = 4(\ln(x))^2 \). This manipulation cleverly isolates \( t \) and translates the equation into a form that's easy to comprehend.

Whenever managing equations, keep in mind to perform the same operation on both sides of the equation, preserving the equality. This is key in maintaining accurate results and arriving at the correct solution in algebra.