Math Problem Statement

The following table gives the U.S. national debt for selected years from 1900 to 2013. Year U.S. Debt ($ billions) Year U.S. Debt ($ billions) 1900 1 1990 3233 1910 1 1996 5225 1920 24 2000 5674 1930 16 2005 7933 1940 43 2009 11,957 1945 259 2010 13,529 1955 273 2011 15,476 1965 314 2012 16,067 1975 533 2013 16,856 1985 1823 (a) Using a function of the form y = a*b^x, with x = 0 in 1900 and y equal to the national debt in billions, model the data. (Round your coefficients to four decimal places.) y =
Incorrect: Your answer is incorrect.

(b) Use the model to predict the debt in 2024. (Round your answer to the nearest billion.) $
Incorrect: Your answer is incorrect. billion

(c) Predict the year in which the debt will be $99 trillion ($99,000 billion).

Solution

Let's break down the problem and find the correct answers step-by-step:

Part (a): Finding the exponential model y=abxy = a \cdot b^x

Given the U.S. debt data, we need to model the debt as an exponential function of the form:

y=abxy = a \cdot b^x

where:

  • yy is the debt in billions,
  • xx is the number of years since 1900 (e.g., x=0x = 0 in 1900, x=10x = 10 in 1910, etc.),
  • aa and bb are constants to be determined.

We can use two points from the data to estimate aa and bb. Let's choose two points:

  • (x1,y1)=(0,1)(x_1, y_1) = (0, 1) in 1900 (since x=0x = 0 in 1900 and the debt was 1 billion dollars),
  • (x2,y2)=(109,16,856)(x_2, y_2) = (109, 16,856) in 2013 (since x=109x = 109 in 2013 and the debt was 16,856 billion dollars).

Using these two points, we can substitute them into the model equation y=abxy = a \cdot b^x.

  1. For 1900 (x=0x = 0, y=1y = 1): 1=ab0a=11 = a \cdot b^0 \Rightarrow a = 1 So, the model becomes y=bxy = b^x.

  2. For 2013 (x=109x = 109, y=16,856y = 16,856): 16,856=1b10916,856 = 1 \cdot b^{109} Solving for bb: b109=16,856b^{109} = 16,856 Taking the 109th root of both sides: b=16,856109b = \sqrt[109]{16,856} Using a calculator, we get: b1.0757b \approx 1.0757

Thus, the exponential model is: y=1.0757xy = 1.0757^x

Part (b): Predict the debt in 2024

In 2024, x=124x = 124 (since x=124x = 124 in 2024 when starting from 1900). Using the model y=1.0757xy = 1.0757^x, we can predict the debt:

y=1.0757124y = 1.0757^{124}

Using a calculator, we get:

y27,724.58billion dollarsy \approx 27,724.58 \, \text{billion dollars}

Rounding to the nearest billion, the predicted debt in 2024 is:

27,725billion dollars\boxed{27,725} \, \text{billion dollars}

Part (c): Predict the year in which the debt will be $99 trillion ($99,000 billion)

We need to find the year xx when the debt will be 99,000 billion dollars. Using the model y=1.0757xy = 1.0757^x, set y=99,000y = 99,000 and solve for xx:

99,000=1.0757x99,000 = 1.0757^x

Take the natural logarithm of both sides:

ln(99,000)=xln(1.0757)\ln(99,000) = x \ln(1.0757)

Solving for xx:

x=ln(99,000)ln(1.0757)x = \frac{\ln(99,000)}{\ln(1.0757)}

Using a calculator:

x11.5020.07302157.48x \approx \frac{11.502}{0.07302} \approx 157.48

Since x=0x = 0 in 1900, the predicted year is:

1900+157.4820571900 + 157.48 \approx 2057

Thus, the debt will reach $99 trillion around the year:

2057\boxed{2057}


Let me know if you'd like further details or explanations!

Related Questions:

  1. What is the general form of an exponential growth model?
  2. How do you interpret the constants aa and bb in exponential models?
  3. How do you solve an exponential equation using logarithms?
  4. How would the debt growth model change if we included more complex factors like inflation?
  5. Can this model be used indefinitely, or does it have limitations?

Tip:

Exponential models are often used to predict population growth, financial investments, and more, but they tend to overshoot predictions in the long term because they don't account for limiting factors. Always evaluate their validity over the range you're interested in.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Algebra
Logarithms

Formulas

y = a*b^x (exponential model)
b = (y2 / y1)^(1/(x2-x1))
x = (ln(y) - ln(a)) / ln(b)

Theorems

Exponential Growth Model
Properties of Logarithms

Suitable Grade Level

Grades 10-12