Question

Use \(y=y_{0} e^{k t}\). The effect of advertising decays exponentially. If \(40 \%\) of the population remembers a new product after 3 days, how long will \(20 \%\) remember it?

Short answer

It takes approximately 7.40 days for 20% of the population to remember the product.

Step-by-step solution

Understand the Formula

The formula given is \( y = y_{0} e^{kt} \), which describes exponential decay. Here, \( y \) is the amount remembered at time \( t \), \( y_0 \) is the initial amount remembered, and \( k \) is the decay constant. Our task is to find the time taken for a certain amount of population to remember the product.

Set Up the Known Values for First Scenario

Initially, \(40\%\) of the population remembers the product after 3 days. Using the formula: \[ 0.4 = y_{0} e^{3k} \].

Solve for \( y_0 \)

We assume everyone remembered it initially, so \( y_0 = 1 \). Thus, the equation simplifies to:\[ 0.4 = 1 imes e^{3k} \] which means\[ e^{3k} = 0.4 \].

Solve for \( k \)

Take the natural logarithm of both sides: \(3k = ext{ln}(0.4)\) Then, solve for \( k \):\(k = \frac{\text{ln}(0.4)}{3}\).

Set Up the Equation for the Second Scenario

Now, we know \( y_{0} = 1 \) and we need to find \( t \) for \( 20\% \) remembering. Set up the equation as:\[ 0.2 = 1 imes e^{kt} \].

Solve for \( t \)

Solve \[ e^{kt} = 0.2 \],take the natural logarithm:\( kt = ext{ln}(0.2) \)Use the value of \( k \) from Step 4:\( \frac{\text{ln}(0.2)}{k} = t \).

Calculate \( t \)

Substitute \( k = \frac{\text{ln}(0.4)}{3} \) into the equation:\[t = \frac{\text{ln}(0.2)}{\frac{\text{ln}(0.4)}{3}}\]Calculate:\[t = \frac{3 imes ext{ln}(0.2)}{ ext{ln}(0.4)} \approx 7.40 ext{ days}\]

Key concepts

Advertising Effectiveness

Understanding advertising effectiveness is crucial in determining how well a campaign reaches and influences its audience. When we talk about effectiveness in this context, we often refer to how much of the target population retains knowledge of or interest in a product over time. Marketing teams aim to create ads that leave a lasting memory, but over time, the effect naturally diminishes.
If an ad initially impacts 100% of its viewers, this percentage will typically decrease in the following days or weeks unless continuously reinforced. This reduction in recall or awareness is known as exponential decay in advertising effectiveness. Different factors can influence this decay, such as the quality of the ad and the frequency with which the product or message is encountered.
In summary, understanding and calculating advertising effectiveness helps marketers decide when and how to re-engage their audience to maintain or boost recall. Calculating the time it takes for viewers to forget about a product is just as important as knowing the percentage that initially remembers it.

Exponential Functions

Exponential functions describe processes that change at rates proportional to their current value. These functions are expressed as \(y = y_{0} e^{kt}\), where:\

• \(y\) is the amount of whatever is being measured (memory, population, radioactivity) at a certain time \(t\).
• \(y_0\) is the initial value or starting point. In advertising, \(y_0\) could represent the percentage of people who initially remember a product.
• \(e\) is the base of the natural logarithm, a constant approximately equal to 2.718.
• \(k\) is the decay (or growth) constant, which dictates how quickly or slowly the value changes.

In advertising, these functions help model and predict how awareness diminishes after initial exposure. The function reflects real-world scenarios where repeated exposures boost remembrance, which follows a diminishing return pattern over time.
By using exponential functions, marketers can quantitatively analyze the effectiveness of various advertising strategies and timeliness. This allows for more strategic budget allocation and timing for advertising investments.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a mathematical function that helps in solving equations involving exponential growth or decay. It is the opposite, or inverse, of the exponential function.
When dealing with exponential decay, such as the decrease in advertising effectiveness, taking the natural logarithm can simplify the process of finding unknowns such as the time \(t\) or the decay constant \(k\). For instance, if \(e^{3k} = 0.4\), taking the natural log of both sides translates it to \(3k = \ln(0.4)\), which is much easier to solve for \(k\).
The natural logarithm is particularly useful for converting problems from multiplicative to additive, making them far simpler to handle and interpret in practical scenarios. It provides the tool needed to break down the exponential equations into linear forms, which are easier to manage and solve in everyday business helps, including the calculation of when to renew advertising efforts.