Question

The number \(A\) of varieties of native prairie grasses per acre within a farming region is approximated by the model \(A=10.5 \cdot 10^{0.04 x}, \quad 0 \leq x \leq 24\) where \(x\) is the number of months since the farming region was plowed. Use this model to approximate the number of months since the region was plowed using a test acre for which \(A=70\)

Short answer

The number of months since the farming region was plowed when \(A= 70\) is approximately 41.8 months.

Step-by-step solution

Set up the Equation

Given the equation \(A=10.5 \cdot 10^{0.04 x}\) and the number of varieties \(A=70\), start by setting up your primary equation, which will be: \(70 = 10.5 \cdot 10^{0.04 x}\).

Isolate the Exponential

Divide both sides of the equation by 10.5, allowing you to isolate the exponential on the right-hand side. This will yield: \(70 / 10.5 = 10^{0.04 x}\), which simplifies to \(6.666666667 = 10^{0.04 x}\).

Transform Exponential Form into Logarithmic Form

Transform the equation into logarithmic form. Remember the base of the exponent becomes the base of the log. This appears like: \(\log_{10}(6.666666667) = 0.04x\). In base 10, this simplifies to \(\log(6.666666667) = 0.04x\).

Solve for x

Now, we need to isolate \(x\). We do this by dividing both sides by 0.04, this gives us \(x = \log(6.666666667) / 0.04\). Finally, we use a calculator to compute the value of \(x\).

Verify Solution

It's important to verify our solution by plugging \(x\) back into the original equation to ensure it gives us \(A= 70\).

Key concepts

Logarithmic Functions

Logarithmic functions are crucial when dealing with exponential growth or decay in algebra problems. In simple terms, a logarithm is the inverse of an exponential function. This means if you have an equation like \(b^y = x\), the logarithmic form is \(y = \log_b(x)\). This property is particularly useful in solving equations involving exponents.

In the problem, we used the logarithm to solve for \(x\) in the equation \(70 = 10.5 \cdot 10^{0.04x}\). By isolating \(10^{0.04x}\) and converting to logarithmic form, \(\log_{10}(6.67) = 0.04x\), we could easily solve for \(x\).

The base 10 logarithm, also known as the common logarithm, is often used in such calculations because it relates directly to exponential functions with base 10. It allows us to convert multiplicative relationships into additive ones, making complex equations easier to handle.

Modeling with Functions

Modeling with functions involves using mathematical equations to represent real-world scenarios. This allows us to make predictions or analyze situations over time. In our example, the equation \(A = 10.5 \cdot 10^{0.04x}\) models the growth of prairie grasses.

Using models helps us understand how variables interact. By knowing the starting conditions and the growth rate, we can predict changes over time, as seen in our task of estimating how many months since plowing by knowing the number of grass varieties.

• The base value, \(10.5\), represents the approximate initial number of varieties per acre.
• The exponential component \(10^{0.04x}\) reflects growth over time, where \(x\) is the number of months.

This type of model is common in ecological and biological studies, where growth is not linear but exponential.

Problem Solving in Algebra

Problem-solving in algebra often involves a series of logical steps to isolate and solve for the unknown variable. This process starts with understanding the given equation and what needs to be solved.

In the grassroots growth problem, the unknown was \(x\), the number of months since plowing. We tackled the problem by:

• Setting up the equation with \(A = 70\).
• Isolating the exponential term by dividing both sides by 10.5.
• Converting to a logarithmic equation to solve for \(x\).
• Verifying the solution to ensure accuracy.

This structured approach helps tackle even complex algebra problems by breaking them into manageable pieces. Consistently applying this method aids in building problem-solving skills that are useful in various mathematical and real-life situations.