# Math Help - explorations

1. ## explorations

Assume that there were 1 million bacteria on an apple at noon on Monday and that a bacterium doubles every hour on average. Let f(t)=M represent the number of bacteria(in millions) on an apple at t hours since noon on Monday. An equation for f is f(t)=2^t.

1. Explain why f^-1(M)=log2(M)

2. Find log2(8). What does your result mean in terms of the bacterium?

2. Originally Posted by getnaphd
Assume that there were 1 million bacteria on an apple at noon on Monday and that a bacterium doubles every hour on average. Let f(t)=M represent the number of bacteria(in millions) on an apple at t hours since noon on Monday. An equation for f is f(t)=2^t.

1. Explain why f^-1(M)=log2(M)
For the inverse function, the output becomes the input. thus, to find the inverse function, we switch the output and input, and solve for the new input (which is the old output).

Let $f(M) = y$

Thus we have $y = 2^M$

For $f^{-1}(M)$:

$M = 2^y$ ........now we solve for $y$

Take log to the base 2 of both sides, we obtain

$\log_2 M = \log_2 2^y$

$\Rightarrow \log_2 M = y$

$\Rightarrow f^{-1}(t) = \log_2 M$

as desired

2. Find log2(8).
$\log_2 8 = 3$ ...not much thinking required there. remember what the definition of a logarithm is, i've stated it several times on this forum, do a search for it if you wish. for the sake of completeness we will show it more explicitly

Let $\log_2 8 = x$

$\Rightarrow 2^x = 8$

$\Rightarrow 2^x = 2^3$

Equating coefficients we get:

$x = 3$

$\Rightarrow \log_2 8 = 3$

What does your result mean in terms of the bacterium?
after 3 hours we have 8 million bacteria.

do you see why?

3. Originally Posted by getnaphd
Assume that there were 1 million bacteria on an apple at noon on Monday and that a bacterium doubles every hour on average. Let f(t)=M represent the number of bacteria(in millions) on an apple at t hours since noon on Monday. An equation for f is f(t)=2^t.

1. Explain why f^-1(M)=log2(M)
Let

$y=f^{-1}(M)$,

then by definition $f(y)=M$, so we have:

$f(y)=2^y=M$

So we need to solve:

$M=2^y$,

for $y$, we do this by taking logs, and the most convienient base in this case is 2, hence:

$\log_2(M)=\log_2(2^y)=y\log_2(2)=y$

Therefore $y=\log_2(M)$, which means:

$f^{-1}(M)=\log_2(M)$

as required.

2. Find log2(8). What does your result mean in terms of the bacterium?
$8=2^3$ so $\log_2(8)=\log_2(2^3)=3 \log_2(2)=3$

RonL

4. Originally Posted by getnaphd
Assume that there were 1 million bacteria on an apple at noon on Monday and that a bacterium doubles every hour on average. Let f(t)=M represent the number of bacteria(in millions) on an apple at t hours since noon on Monday. An equation for f is f(t)=2^t.

1. Explain why f^-1(M)=log2(M)

2. Find log2(8). What does your result mean in terms of the bacterium?
1) one if f(t)=m then all you have is a line parrallel to the x-axis. So it would be better to say that f(t)=M=2^t in millions of bacteria per hour. if this is true then M=2^t and

log2(M)=log2(2^t)

log2(M)=t*log2(2)

log2(M)=t

honestly this is probably what you were looking for. and I realy can't dertmine what you were trying to get at in the first part of the equation especaily since f is not a varible that was any of the equations. mabe you were trying to say F(m^-1) or f(M)^-1 were you?

at least I can answer the final question

2) log2 (8)

2*2*2=2^3=8

log2(2^3)= 3 log2(2)= 3(1)= 3

this means that it takes 3 hours to make 8 million bacterium

5. wow all three of us were working on the same problem but hey thanks for helping me realize that he was trying to invert the function. wow I must be realy tired.