Problem 41 - Precalculus: Mathematics for Calculus

Question

Follow the hints and solve the systems. $$\left\\{\begin{aligned} \log x+\log y &=\frac{3}{2} \quad[\text {Hint: Add the equations.}] \\ 2 \log x-\log y &=0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

\( x = \sqrt{10}, y = 10 \)

Step 1: Apply Logarithmic Properties

Use the logarithmic property that states \( \log a + \log b = \log(ab) \). Thus, for the first equation, \( \log x + \log y = \frac{3}{2} \) becomes \( \log(xy) = \frac{3}{2} \).

Step 2: Simplify the Second Equation

The second equation is \( 2 \log x - \log y = 0 \). Rearrange it to \( \log x^2 = \log y \). This implies \( x^2 = y \) by properties of logarithms.

Step 3: Replace \( y \) with \( x^2 \) in the First Equation

Substitute \( y = x^2 \) into the first equation \( \log(xy) = \frac{3}{2} \). Therefore, \( \log(x \cdot x^2) = \frac{3}{2} \) or \( \log(x^3) = \frac{3}{2} \).

Step 4: Solve for \( x \)

Since \( \log(x^3) = \frac{3}{2} \), this implies \( x^3 = 10^{3/2} \). Therefore, \( x = \sqrt[3]{10^{3/2}} = 10^{1/2} = \sqrt{10} \).

Step 5: Solve for \( y \)

Given that \( y = x^2 \) and we have \( x = \sqrt{10} \), substitute to find \( y = (\sqrt{10})^2 = 10 \).

Key Concepts

Logarithmic Properties

Logarithmic properties are immensely helpful in simplifying complex expressions that involve logarithms. They allow us to combine and break apart logs in a way that's more manageable. In this exercise, we utilized the property:

• The product property: \( \log a + \log b = \log(ab) \).

This property lets us express the sum of two logarithms as the logarithm of a single product, simplifying calculations greatly.
For example, the step that converts \( \log x + \log y = \frac{3}{2} \) to \( \log(xy) = \frac{3}{2} \) is founded on this property. By transforming the equation, it becomes easier to work with and helps in solving the problem efficiently by reducing the number of terms involved.
Remembering these properties can speed up calculations and simplify solving logarithmic equations, which is incredibly beneficial in handling more intricate systems of equations, like the one in this exercise.

Systems of Equations

A system of equations consists of two or more equations with the same set of unknowns. Solving these systems involves finding the values of the variables that satisfy all the equations simultaneously. In the context of logarithmic equations, this can often involve substitution or elimination methods.
In our example, we have two equations:

• \( \log x + \log y = \frac{3}{2} \)
• \( 2 \log x - \log y = 0 \)

To solve them, we first simplified one equation using logarithmic properties, allowing us to express one variable in terms of the other, specifically \( y = x^2 \).
Substitution involves taking this expression for \( y \) and replacing it in another equation to solve for \( x \). Once we find \( x \), we substitute it back to find \( y \). It simplifies the process significantly, as it transforms the system into simpler parts that are easier to solve individually.

Solving Exponential Equations

When dealing with exponential equations, the aim is often to solve for the unknown variable within the exponent. The logarithmic approach is very effective in these situations because it can convert multiplicative relationships into additive ones, which are simpler to solve.
In this exercise, after simplifying the equations, we were left with \( \log(x^3) = \frac{3}{2} \). This can be rewritten as an exponential equation:

• \( x^3 = 10^{3/2} \)

To find \( x \), we employ the property of logarithms that translates the expression back into an exponential form, allowing for straightforward computation using known values.
Here, taking the cube root, or raising both sides to the power of \( \frac{1}{3} \), gives us \( x = 10^{1/2} \), which simplifies to \( \sqrt{10} \). Hence, we solved the exponential equation by converting the logarithmic relationship into something familiar and easy to compute.
This method shows how logarithmic and exponential equations are interconnected and can be used to solve problems systematically, leveraging their properties to simplify and solve effectively.