Module B Outcome 4&6 Logarithms and Inverses

Jaime decides to invest her money in a bank that advertises 5.5% annual interest compounded monthly. She decides her initial investment will be $1000. Check all the true statements that apply.

• The function that models the growth of Jaimes money is

  $f\left(x\right)=1000\left(1+\frac{.055}{12}\right)^{12x}$ , such that f(x) is the amount of money after a given number of x years.

• The function that models the growth of Jaimes money is $f\left(x\right)=1000\left(1+\frac{5.5}{12}\right)^{12x}$ , such that f(x) is the amount of money after a given number of x years.

• The annual yield of of Jaimes account is 5.64%

• The annual yield if Jaimes account is 5.5%

• $f\left(4\right)=1315.70\$ Indicates that Jaimes account will have $1315.70 after 4 years.

Bill opens an online savings account through Goldman Sach's Marcus. He determines a formula that can be used to predict his accounts growth. Check all that apply regarding Bills account.

 $m\left(x\right)=\ 5000\left(1+\frac{.035}{12}\right)^{12x}$  

• Bill will have $5,954.71 in his account after 5 years

• The inverse of Bills function in  $m^{-1}\left(x\right)=\frac{\log_{\left(1+\frac{.035}{12}\right)}\left(\frac{x}{5000}\right)}{12}$  

• The invers of Bills function is approximately  $m^{-1}\left(x\right)=\frac{1}{12}\log_{\left(1.0029\right)}\left(\frac{x}{5000}\right)$  

•  $m^{-1}\left(6385.83\right)=7$  Indicates that bill will have a total of $6,385.83 after 7 years

Given the function

 $g\left(x\right)=\ 2.225\left(3^x\right)+11$ . Check all that apply

• Given the function $h\left(x\right)=\ 71.25$  , g(x) = h(x) when x = 3

• The Inverse of g(x) is  $g^{-1}\left(x\right)=\ \log_3\left(\frac{x+11}{2.225}\right)$  

• The Inverse of g(x) is  $g^{-1}\left(x\right)=\log_3\left(\frac{x-11}{2.225}\right)$  

• The inverse of g(x) is  $g^{-1}\left(x\right)=\ \log_{\left(\frac{x-11}{2.225}\right)}3$  

 $\log_u125=3$  

Convert the following Logarithmic statement to an exponential statement.

•  $u^3=125$  

•  $3^u=125$  

•  $u^{125}=3$  

•  $3^{125}=u$  

Convert the following exponential equation to a logarithm
 $4^x=18$  

•  $\log_418=x$  

•  $\log_{18}4=x$  

•  $\log_x18=4$  

•  $\log_4x=18$  

Given
 $j\left(x\right)=-2\left|x-3\right|+5$  

Check all that apply.

• All of the transformations that have taken place are a translation right 3 and up 5. The vertical stretch is 2 and a reflection across the x-axis

•  $2x-1\ \ \ \ \ \ \ \ \ \ x\le3$  
   $-2x\ +11\ \ \ \ \ \ \ \ \ \ x>3$  
  

• 
   $2x+2\ \ \ \ \ \ \ x\le3$  
   $-2x+11\ \ \ \ \ \ \ x<3$  

• All of the transformations that have taken place are a translation Left 3 and up 5. The horizontal compression is 2 and a reflection across the x-axis

Given the function
 $g\left(x\right)=\ -3x^2+12x-7$ 

Check all that applies 

• The inverse for the domain restriction  $x\ge2$  is  $g^{-1}\left(x\right)=\sqrt{-\frac{1}{3}\left(x-5\right)}+2$  

• The Inverse for the domain restriction  $x\ge2$  is  $g^{-1}\left(x\right)=\sqrt{\frac{\left(x-5\right)}{3}}+2$  

• Given the function  $t\left(x\right)=\ \frac{1}{3}x+1$  ,  $g\left(x\right)>t\left(x\right)$  on the interval  $\left(.889,\ 3\right)$  

• Given the function  $t\left(x\right)=\ \frac{1}{3}x+1$  ,  $g\left(x\right)>t\left(x\right)$  on the interval  $\left(-\infty,\ .889\right)\ U\ \left(3,\ \infty\right)$  

Given the exponential function
 $j\left(x\right)=\ 100\left(\frac{4}{3}\right)^{\frac{2}{3}x}$  

Check all that apply.

• The exponential function is decay

• The exponential equation is growth

• The percent of growth is 21.1% increase

• The percent of growth is 33.33% increase

• The function can be rewritten as approx.  $j\left(x\right)=133^{\frac{2}{3}x}$  

• The function is decaying and has a percent of decrease of 38.1%

• The function is decaying and has a percent of decrease of 33.3%

A scientist is studying the growth of bacteria. The scientist starts the study with 100 bacteria cells. An hour later the bacteria had multiplied to a total of 112. Check all that apply.

• 
  The function that models the growth of bacteria based on hours is
  $b\left(h\right)=100\left(1.12\right)^h$

  
  

• The function that models the growth of the bacteria based on minutes is $b\left(m\right)=100\left(1.12\right)^{\frac{1}{60}m}$

• The function that models the growth of the bacteria based on days is $b\left(d\right)=100\left(1.12\right)^{24d}$

• The daily growth rate is $1,417.9\%$ increase