Problem 11 - Algebra 2

Question

Find the value of \(y\) for the given value of \(x .\) $$ y=0.05-10 \ln x, \text { for } x=0.09 $$

Step-by-Step Solution

Answer

Upon executing the operations, you will find that \(y = 23.25309708\).

Step 1: Substitution

Substitute \(x = 0.09\) into the given equation. We get: \\(y = 0.05 - 10 \ln(0.09)\)

Step 2: Calculation

Now, calculate the natural logarithm of \(0.09\) and multiply by \(10\). Then subtract the result from \(0.05\). Remember that the ln function refers to the natural logarithm, which is log base \(e\). In most scientific calculators or software, you can directly compute \(\ln(0.09)\).

Step 3: Simplification

After calculating the above expression, you will obtain the value of \(y\).

Key Concepts

Logarithmic Functions

Logarithmic functions are an essential part of mathematics, particularly in the fields of algebra and calculus. These functions are used to solve equations where the variable is an exponent. The natural logarithm, denoted as \(\ln\), is a type of logarithm that uses the base \(e\), which is approximately 2.718. This base is unique as it arises naturally in various mathematical contexts, including compound interest calculations and growth models in sciences.
When you see \(\ln(x)\), it essentially means, "What power should \(e\) be raised to, to get \(x\)?". Consequently, calculating \(\ln(0.09)\) asks, "What power of \(e\) gives us 0.09?". This logarithmic function is often used in financial calculations, scientific processes, and, as seen in the problem, to solve equations involving growth or decay.
For our exercise, understanding the conversion of the real-world number \(x\) into its logarithmic form via \(\ln\) helps determine the influence of \(x\) on another variable \(y\). Calculating the exact logarithmic value requires computational tools like calculators or software since it often results in non-intuitive fractions or decimals.

Exponential Functions

Exponential functions play a reciprocal role to logarithmic functions. While logarithms help solve for an unknown exponent, exponential functions grow or decay at a steady rate and have the general form \(y = a \cdot e^{bx}\). Here, \(a\) and \(b\) are constants, and \(e\) is the base of natural logarithms. They are characterized by such properties as constant proportional growth or decay, making them very useful in population dynamics, radioactive decay, and financial forecasting.
When we are given an equation like \(y = 0.05 - 10 \ln(x)\), although it directly involves a logarithmic expression, there's an implicit relationship that reflects exponential behavior due to the existence of \(\ln\).
Understanding exponential functions, you get to reverse-engineer processes described logarithmically, allowing for a better grasp of growth rates and their natural and interlinked dependence on logarithmic functions. For learners, seeing how exponential and logarithmic functions complement each other aids in the comprehension of complex mathematical relationships.

Algebraic Substitution

Algebraic substitution is a basic yet powerful technique used in algebra to simplify and solve equations. It involves replacing a variable with a given number or another expression, making calculations more manageable. In the context of the problem, substituting \(x = 0.09\) into the function \(y = 0.05 - 10 \ln(x)\) allows us to compute \(y\) when \(x\) is known.

• This process allows for step-by-step evaluation - firstly substituting the specific value of \(x\), then calculating \(\ln(0.09)\), and subsequently performing operations to find \(y\).
• Substitution is particularly useful in verifying solutions and bridging between variables and their numerical values. This systematic approach aids in maintaining clarity throughout problem-solving.

For this exercise, accurate substitution followed by careful computation of the natural logarithm sets the stage for evaluating the effect of \(x\)'s value on the result of \(y\). Learning to perform substitution effectively is an indispensable skill for handling more complex algebraic manipulations.