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# Module 23

### Chapter 12.3

Question | Answer |
---|---|

Find f(x) by finding three points on the line f(x)=2^x. | Plug in some points from a graph. Remember to correctly plot points on graph and connect the points. X: -2 , 0 , 2 f(x): (1/4) , 1 , 4 |

Find y for the exponential function (1/2)^x. | Plug in some points from a graph. X: -1 , 0 , 1 Y: 2 , 1 , (1/2) |

Find 4 points on the line of the exponential function f(x)=3^x+2. | Plug in plots from graph. Choose smaller numbers so you can plot the points easier. X: -3 , -2 , -1 , 0 f(x): (1/3) , 1 , 3 , 9 |

Solve for x. 2^x=16 | Change base of 16 so that it matches 2. 1. 2^x=2^4 2. x=4 |

Solve for x. 9^x=27 | Change base of 27 so that it matches 9. Both bases are powers of 3. 1. (3^2)^x=3^3 2. Solve for x. - 2^x=3 - (2^x)/2=3/2 - x=3/2 |

Find the amount owed at the end of 5 years if $1600 is loaned at a rate of %9 compounded monthly. | P= $1600 r= 0.09 n= 12 t= 5 Plug in the values into the equation. A= $2505.09 |

A type of uranium has daily radioactive decay of 0.4%. If 40 pounds of this uranium is available today, find how much will still remain after 40 days. Use y=40(2.7)^-0.004t | t= 40 Plug in t value. y= 34.1 |

The equation y=84,943(1.095)^x models the number of college students who study abroad each year from 1995 through 2006. In the equation, x represents the number of years after 1995. Find the # of students studying abroad in (a) 2001 and (b) 2023. | Plug in x for the # of years after 1995. (a) 84,943(1.095)^6=146,424. (b) 84,943(1.095)^28=1,078,259 |

The equation y=114.65(1.114)^x gives the number of cellular phone users y (in millions) in a country for the years 2002 through 2009. In this equation, x=1 represents 2003. Find out how many users there will be in 2012. | Plug in x for the # of years up to 2012. 114.65(1.114)^10=337.5 |

Solve for x. 4^x+3=8^x | 1. (2^2)^x+3=(2^3)^x 2. (2x+6)-2x=3x-2x 3. 6=x |

Find f(x) by finding four points on the line f(x)3^x. | X: -2 , -1 , 3 ,2 f(x): (1/9) , (1/3) , 27 , 9 |

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