Problem 18 - Algebra 2

Question

Solve each equation. Check your answers. $$ \ln (t-1)^{2}=3 $$

Step-by-Step Solution

Answer

The solutions to the given equation are \( t = 1 + \sqrt{e^{3}} \) and \( t = 1 - \sqrt{e^{3}} \), subject to the condition that none of the solutions return a negative logarithmic argument when substituted into the original equation.

Step 1: Remove the Logarithm

Use properties of logarithms to undo the logarithm and isolate the quadratic equation. Because \( \ln a = b \) can be written as \( e^{b} = a \), the equation can be rewritten as \[ (t-1)^{2} = e^{3} \]

Step 2: Solve for the Quadratic Equation

Taking the square root of both sides, remember there should be a positive and negative solution because it is a quadratic equation in the form \( (a-b)^{2} = c \). Thus we have \[ t-1 = \pm \sqrt{e^{3}} \] or equivalently \[ t = 1 \pm \sqrt{e^{3}} \]

Step 3: Check Your Answer

Both answers need to be checked in the original logarithmic equation. Replacing \( t \) in the original equation with \( 1 + \sqrt{e^{3}} \) and \( 1 - \sqrt{e^{3}} \) respectively, if none of the answers cause the argument of the logarithm to be negative (as the logarithm of a negative number is undefined), then both are valid answers.

Key Concepts

Properties of Logarithms

Logarithms have fascinating properties that make them magical tools in solving equations. Among these properties are:

• Product Rule: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• Quotient Rule: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• Power Rule: \( \log_b(M^p) = p \cdot \log_b(M) \)

In the given exercise, we use the Power Rule to simplify \( \ln (t-1)^2 = 3 \). This property allows us to transform the logarithmic equation into a form that is easier to manipulate. The inverse relationship between logarithms and exponents is utilized by rewriting \( \ln a = b \) as \( e^b = a \). This step is crucial as it sets the stage for solving the equation by removing the logarithmic barrier, which is often the most intimidating part of such problems. Once the logarithm is dealt away with, we are left with a manageable quadratic equation.

Quadratic Equations

Quadratic equations have the standard form \( ax^2 + bx + c = 0 \). Solving them involves finding values of \( x \) that make the equation true. In our problem, the equation \( (t-1)^2 = e^3 \) resembles a simple quadratic, focusing on the expression \( (a-b)^2 = c \).

When you find solutions to such equations, it's important to consider both positive and negative solutions due to the squared nature. By taking the square root of both sides, we get:

• \( t - 1 = \sqrt{e^{3}} \)
• \( t - 1 = -\sqrt{e^{3}} \)

Thus, providing two potential solutions:

• \( t = 1 + \sqrt{e^{3}} \)
• \( t = 1 - \sqrt{e^{3}} \)

Each solution is derived from acknowledging \( \pm \sqrt{e^{3}} \), highlighting the dual nature inherent in quadratic equations.

Checking Solutions

After solving for \( t \), it's essential to verify whether these solutions satisfy the initial problem conditions. This process is crucial as it helps avoid errors such as incorrect mathematical operations or invalid domain values.

To check solutions, substitute each value back into the original equation \( \ln (t-1)^2 = 3 \) and ensure that:

• The logarithm does not have a negative argument.
• The result equals the original equation condition.

For example, using \( t = 1 + \sqrt{e^{3}} \), we plug back into the logarithmic equation. The expression remains valid because \( \sqrt{e^3} \) is a positive value, maintaining the logarithm's domain of non-negative numbers. Following the same steps with \( t = 1 - \sqrt{e^{3}} \), if both solutions satisfy the original equation, they are verified as correct.

Exponential Equations

Exponential equations involve expressions where variables are located in an exponent, like \( b^x = a \). They are the inverse operations of logarithms.

In solving our problem, we transform the logarithmic equation into an exponential form: \( (t-1)^2 = e^3 \). This tackled the logarithmic aspect and shifted the problem to a realm where square roots and quadratics offer resolution paths.

• Step from \( \ln (t-1)^2 = 3 \) to \( e^3 = (t-1)^2 \) epitomizes leveraging the relationship between logs and exponents.
• By converting it into an exponential equation, it becomes more straightforward, unveiling solutions through familiar arithmetic operations like square roots.

Once converted, solving proceeds with simpler, intuitive processes, leveraging exponent rules and knowledge from basic algebra.