Problem 16 - Algebra 2

Question

Solve each equation or inequality. Round to the nearest ten-thousandth. \(\ln x^{2}=9\)

Step-by-Step Solution

Answer

The solutions are approximately \( x \approx 90.0232 \) and \( x \approx -90.0232 \).

Step 1: Convert to Exponential Form

To solve the equation \( \ln x^{2} = 9 \), recognize that the natural logarithm \( \ln \) implies that we are using base \( e \). Thus, we can rewrite the equation in exponential form as \( e^{9} = x^{2} \).

Step 2: Solve for \(x^2\)

The equation from Step 1, \( x^{2} = e^{9} \), allows us to solve for \( x \) by taking the square root on both sides. We will have: \( x = \pm \sqrt{e^{9}} \).

Step 3: Calculate \( e^{9} \)

Calculate \( e^{9} \) using a calculator. You will find that \( e^{9} \approx 8103.0839 \).

Step 4: Determine \( x \)

Use the approximate value of \( e^{9} \) to find \( x \). Thus, \( x = \pm \sqrt{8103.0839} \). Calculate the square root to find \( x \approx \pm 90.0232 \).

Step 5: Round to the Nearest Ten-Thousandth

Round the value of \( x \) to the nearest ten-thousandth. Both values, \( x \approx 90.0232 \) and \( x \approx -90.0232 \), are already at this precision.

Key Concepts

Exponential Form

The natural logarithm, denoted as \( \ln \), sets its base as \( e \), which is an irrational number approximately equal to 2.71828. When we express an equation involving \( \ln \) in its exponential form, we're essentially rotating the equation around the base \( e \). For example, the given equation \( \ln x^{2} = 9 \) translates into an exponential expression by restructuring it into \( e^{9} = x^{2} \). This means that we are finding the power of \( e \) that equals \( x^{2} \). Converting logarithmic equations to exponential form often makes finding solutions more straightforward and highlights the power relationship between the numbers.

Square Root

After expressing our equation in exponential form, we arrive at \( x^{2} = e^{9} \). This requires us to solve for \( x \), which involves the mathematical operation of taking a square root. The square root of a number \( a \), denoted \( \sqrt{a} \), is the value that, when multiplied by itself, gives \( a \). For our equation, we take the square root of both sides, yielding \( x = \pm \sqrt{e^{9}} \). The "\( \pm \)" symbol indicates that both positive and negative roots need consideration because both values, when squared, satisfy the equation. This dual solution is a critical component when dealing with squared variables in equations.

Approximation

The exponential expression \( e^{9} \) often results in non-integral, unwieldy numbers. This is why approximation is useful, especially when dealing with significant decimals. Using approximation tools, or calculators, we find \( e^{9} \approx 8103.0839 \). Approximating numbers lets us comprehend and compute large values more manageably by adjusting them to more familiar decimal numbers. It’s important to recognize that approximations do not represent exact values, but provide a practical means to perform calculations efficiently.

Rounding Numbers

Rounding numbers helps simplify complex numbers by reducing their decimal places, aiding in easier comprehension and representation. To round a number to the nearest ten-thousandth, you need to check the fifth decimal place. If it is 5 or greater, you increase the decimal place you are rounding to by one; if it's less, you retain the digit. In our exercise, we calculated \( x \approx \pm 90.0232 \), which is already at the required precision. Proper rounding ensures that the result retains adequate accuracy while discarding unnecessary decimal excess, improving readability and usefulness in further calculations.