Problem 52 - Algebra 2

Question

Solve each equation. Round to four decimal places. $$ 3^{y}=\sqrt{2^{y-1}} $$

Step-by-Step Solution

Answer

Using a calculator, the solution is approximately \(y = -3.6974\).

Step 1: Understand the problem

We are given the equation \(3^{y} = \sqrt{2^{y-1}}\). Our goal is to find the value of \(y\) that satisfies this equation and then round the answer to four decimal places.

Step 2: Simplify the square root

First, simplify the square root term. The square root \(\sqrt{2^{y-1}}\) can be rewritten using exponential notation as \((2^{y-1})^{0.5}\). This simplifies to \(2^{0.5(y-1)}\).

Step 3: Equate exponents

After rewriting the right side of the equation, our equation becomes \(3^{y} = 2^{0.5(y-1)}\). We can solve this equation by taking the natural logarithm (ln) of both sides to handle the exponents.

Step 4: Apply logarithms

Applying the natural logarithm to both sides, we get \(\ln(3^{y}) = \ln(2^{0.5(y-1)})\). By the power rule of logarithms, this becomes \(y \cdot \ln(3) = 0.5(y-1) \cdot \ln(2)\).

Step 5: Isolate y

Distribute the 0.5 on the right-hand side: \(y \cdot \ln(3) = 0.5y \cdot \ln(2) - 0.5 \cdot \ln(2)\). Move terms involving \(y\) to one side: \(y \cdot \ln(3) - 0.5y \cdot \ln(2) = -0.5 \cdot \ln(2)\). Factor out \(y\): \(y(\ln(3) - 0.5\ln(2)) = -0.5 \cdot \ln(2)\).

Step 6: Solve for y

Divide both sides by \(\ln(3) - 0.5\ln(2)\) to get \(y = \frac{-0.5 \cdot \ln(2)}{\ln(3) - 0.5 \cdot \ln(2)}\). Calculate this expression using a calculator, keeping precision to four decimal places, to find \(y\).

Step 7: Calculate value

Calculate \(y = \frac{-0.5 \cdot \ln(2)}{\ln(3) - 0.5 \cdot \ln(2)}\) using a calculator. The values for the natural logarithms are approximately \(\ln(2) \approx 0.6931\) and \(\ln(3) \approx 1.0986\). Substitute and calculate to find \(y\).

Key Concepts

Exponential Functions

Exponential functions have equations where variables appear as exponents. For example, in the equation given, \(3^{y}\), the "\(3\)" is a base and the "\(y\)" is an exponent. Exponential functions are used to represent growth or decay scenarios because they model phenomena where growth depends multiplicatively on the current quantity. This is why exponential functions appear often in compound interest, population growth, and radioactive decay problems.
When solving exponential equations, one fundamental step is to make bases the same, if possible, or to use logarithms to manage the exponents.

• When bases on both sides are equal, compare exponents and solve.
• If bases differ, use logarithms which help by "bringing down" exponents.

Once you learn the behavior of exponential functions, solving equations involving them becomes much more intuitive. You'll see their power particularly when combined with logarithmic functions.

Natural Logarithm

A natural logarithm, often denoted as \(\ln x\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to \(2.71828\). Natural logarithms are highly regarded in calculus and mathematical analysis as they appear naturally in various scientific contexts.
A key property of logarithms is they transform multiplication into addition, making them an excellent tool to simplify exponential equations. For instance, the equation \(\ln(3^y) = \ln(2^{0.5(y-1)})\) is simplified using the power rule of logarithms:
- \(\ln(a^b) = b \ln(a)\)
By applying this property, you can "bring down" the exponents; converting a potentially tricky multiplication into a manageable linear scale. This characteristic allows for straightforward algebraic manipulation.

Equation Solving Methods

Equation solving methods can vary depending on the type and complexity of the equation. For the example equation \(3^{y} = \sqrt{2^{y-1}}\), a systematic approach is vital to find the correct solution.
Here are steps often taken to tackle such equations:

• Simplify the equation using known algebraic identities and properties, like converting roots to fractional exponents.
• Use logarithms to linearize exponential relationships, thereby simplifying the comparison between expressions.
• Apply algebraic manipulations like distribution, factoring, and isolating terms to bring the equation to a solvable state.

Using these techniques, an equation becomes simpler to handle. By isolating the variable of interest through strategic algebra and applying the properties of exponential and logarithmic functions, you can solve and find the numerical value of the variable, ensuring you reach an accurate solution rounded to the requisite decimal places.