Problem 17 - Algebra 2 and Trigonometry

Question

In \(11-22,\) solve each equation for \(y\) in terms of \(x\) $$ x=12^{-y} $$

Step-by-Step Solution

Answer

The solution for \(y\) in terms of \(x\) is \(y = -\frac{\ln(x)}{\ln(12)}\).

Step 1: Understand the Problem

We need to solve the equation \(x = 12^{-y}\) for \(y\) in terms of \(x\). This means we'll express \(y\) in an equation where \(x\) is the variable on one side.

Step 2: Rewrite the Equation

To isolate \(y\), we first rewrite the given equation. The equation is \(x = 12^{-y}\). This is an exponential equation with base \(12\) and exponent \(-y\).

Step 3: Apply Logarithms

To solve the equation for \(y\), we apply the logarithm to both sides. This involves taking the natural logarithm (\(\ln\)) for convenience or the common logarithm (log base 10). Here we choose the natural logarithm:\[ \ln(x) = \ln(12^{-y}) \] Using the power rule for logarithms, this can be written as:\[ \ln(x) = -y \cdot \ln(12) \]

Step 4: Isolate y

We solve for \(y\) by dividing both sides of the equation by \(-\ln(12)\):\[ y = -\frac{\ln(x)}{\ln(12)} \] Now \(y\) is expressed in terms of \(x\).

Key Concepts

Natural Logarithm

The natural logarithm is a mathematical function denoted by \( \ln \). It is the inverse operation of exponentiation with the base \( e \), where \( e \) is Euler's number (approximately 2.71828). When you see \( \ln(x) \), it asks "to what power must \( e \) be raised, to equal \( x \)?" This function is a key tool for solving exponential equations.

Using the natural logarithm allows us to manipulate exponents into a linear form that is easier to solve. In the step-by-step solution, we apply \( \ln \) to both sides of the equation \( x = 12^{-y} \). It helps transform the equation into \( \ln(x) = -y \cdot \ln(12) \).

• This utilizes the property of logarithms: \( \ln(a^b) = b \cdot \ln(a) \).
• It's particularly useful when the variable is in the exponent, making it indispensable in exponential equation solving.

Exponential Equation Solving

Exponential equations involve variables as exponents. These types of equations can seem challenging, but with the right techniques, they become manageable. In our equation \( x = 12^{-y} \), our goal is to solve for \( y \).

The first challenge is that \( y \) is in the exponent position, which requires us to "bring it down" to solve the equation. Applying logarithms to both sides helps achieve this by changing the form of the equation.

• We start by taking the natural logarithm on both sides: \( \ln(x) = \ln(12^{-y}) \).
• This changes the equation to: \( \ln(x) = -y \cdot \ln(12) \).

This transformation allows us to apply algebraic techniques to find \( y \). After applying logarithms, you are often left with a linear equation instead of an exponential one, making it easier to isolate the variable and solve.

Isolation of Variables

Isolation of the variable is a critical step in solving equations. In the provided exercise, we aim to express \( y \) solely in terms of \( x \). This means that by manipulating the equation, \( y \) becomes the subject, with only \( x \) and constants on the other side.

In the equation \( \ln(x) = -y \cdot \ln(12) \), we isolate \( y \) by dividing both sides by \( -\ln(12) \):

• Perform this division: \( y = \frac{-\ln(x)}{\ln(12)} \).
• This final expression reveals \( y \) in the simplest form, related directly to \( x \).

Isolating variables not only simplifies the equation but also provides a clear format to understand the relationship between \( x \) and \( y \). It's a fundamental technique used across various types of equations, ensuring clarity and solutions at a glance.