Problem 17 - Calculus

Question

Solve for \(x\) using logs. $$2 x-1=e^{\ln x^{2}}$$

Step-by-Step Solution

Answer

The solution is \(x = 1\).

Step 1: Simplify the Right Side

Start by simplifying the right side of the equation \(e^{\ln x^{2}}\). Since the exponential function \(e\) and the natural log function \(\ln\) are inverse functions, \(e^{\ln x^{2}} = x^{2}\). The equation simplifies to \(2x - 1 = x^{2}\).

Step 2: Rearrange the Equation

Now that we have \(2x - 1 = x^{2}\), move all terms to one side of the equation to set it to zero: \(x^{2} - 2x + 1 = 0\).

Step 3: Recognize the Quadratic Form

Notice that the equation \(x^{2} - 2x + 1 = 0\) is a quadratic equation in the form \(ax^{2} + bx + c = 0\). Here, \(a = 1\), \(b = -2\), and \(c = 1\).

Step 4: Factor the Quadratic

Factor the quadratic equation \(x^{2} - 2x + 1 = 0\). This can be written as \((x-1)^{2} = 0\).

Step 5: Solve for \(x\)

Since \((x-1)^{2} = 0\), the solution is \(x - 1 = 0\). Thus, \(x = 1\).

Key Concepts

Exponential and Logarithm Inverse Properties

The equation given in the problem, \( e^{\ln x^{2}} \), showcases an important property of logarithms and exponential functions: the inverse property. Understanding this property can make solving equations like these much easier.
Exponential and logarithm functions are inverses of each other. This means that they "undo" each other's effects on a number. More specifically, if you have \( e^{\ln a} \), it simplifies to \( a \) because the exponential function \( e \) and the natural logarithm \( \ln \) cancel out. Similarly, \( \ln(e^b) = b\).
This inverse relationship allows for simplification in equations. Applying it to \( e^{\ln x^{2}} \), we see that the expression reduces simply to \( x^{2} \), thus simplifying the problem significantly. Recognizing and applying these inverse rules is crucial when dealing with logarithmic and exponential equations.
When solving logarithmic equations, always keep an eye out for opportunities to apply this property to make your equation more manageable.

Quadratic Equation

A quadratic equation is any equation that can be rearranged into the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These types of equations represent a parabola when graphed in a coordinate plane.
In the context of our exercise, once we simplified the expression \( e^{\ln x^{2}} \) to \( x^{2} \), we rearranged the expression \( 2x - 1 = x^{2} \) to fit this standard quadratic form. The rearranged equation became \( x^{2} - 2x + 1 = 0 \), where \( a = 1 \), \( b = -2 \), and \( c = 1 \).
Handling quadratic equations effectively is essential for solving many algebraic problems. Once in standard form, these equations can be solved using various methods, such as factoring, completing the square, or using the quadratic formula depending on the specific equation at hand.

Factoring Quadratic Equations

Once a quadratic is in the standard form, one efficient technique to find its solutions is factoring. Factoring involves breaking down the equation into simpler binomial factors that, when multiplied together, give the original quadratic.
With \( x^{2} - 2x + 1 = 0 \), the quadratic is already in a form that's straightforward to factor. Notice that it can be expressed as the square of a binomial: \((x-1)^2 = 0\). This is a result of recognizing a perfect square trinomial, where the first and last terms are perfect squares and the middle term is twice the product of their square roots.
Factoring helps us see that \((x-1)^2 = 0\) implies \(x-1 = 0\). Solving this quickly provides the solution \(x = 1\).
Mastering factoring techniques for quadratics can greatly simplify the solving process and is useful in higher-level math studies. Remember, not all quadratics are easily factorable, and in those cases, other methods like completing the square or the quadratic formula are useful alternatives.