1. ## f(x)=b*a^x

Here is a question I want you to look into. I want to see where I have gone wrong.

In 1948 there was 823 cancerpatients. I 1993 there was 5707 cancerpatients. Find the values a and b. Facit: a = 1.045 b = 770

This is what I have done to find these values:

1993-1948= 45 years

823=b*a^0 --> 823/(a^0) = b

5707=b*a^45 --> 5707/(a^45) = b

823/(a^0) = 5707/(a^45) --> (a^45)/(a^0) = 5707/823

a^0 = 1

a = 45sq.root(5707/823) = 1.044

Now I have the value for a. Now for b:

5707/(1.044^45) = 822
eller
823/(1.044^0) = 823

How come b is not the same as the facit? What did I do wrong?

2. Hello, No Logic Sense!

I see absolutely nothing wrong with your work . . .

In 1948 there was 823 cancer patients.
In 1993 there was 5707 cancer patients.
Find the values $a\text{ and }b.$
. . Facit: $a = 1.045\;{\color{red}?}\;\; b = 770\;\;{\color{red}??}$
We have: . $f(t) \:=\:ba^t$

And we have: . $\begin{array}{cc}1948\:(t = 0) & 823 \\ 1993\:(t = 45) & 5707 \end{array}$

$f(0) = 823\quad\Rightarrow\quad ba^0 \:=\:823 \quad\Rightarrow\quad\boxed{ b \:=\:823}$

The function (so far) is: . $f(x) \:=\:823a^t$

$f(45) = 5707\quad\Rightarrow\quad 823a^{45} \:=\:5707\quad\Rightarrow\quad a^{45} \:=\:\frac{5707}{823}$

Then: . $a \:=\:\left(\frac{5707}{823}\right)^{\frac{1}{45}} \:=\:1.043972521 \quad\Rightarrow\quad\boxed{ a \:\approx\:1.044}$

Therefore, the function is: . $\boxed{f(t) \;=\;823(1.044)^t}$

3. My teacher said that it has something to do with drawing the function on a logarithm paper. Apparently if you do that, you will get the b value to 770.

4. Originally Posted by No Logic Sense
My teacher said that it has something to do with drawing the function on a logarithm paper. Apparently if you do that, you will get the b value to 770.
logarithm paper?

5. Originally Posted by colby2152
logarithm paper?
I think he means the log graph sheet, where the x axis is in logarithmic scale and y axis is normal.

6. Most "log graph paper" has a normal x-axis and a logarithmic y-axis.
. . Then an exponential function becomes a line.

We have . $y \;=\;823(1.044)^x$

Take logs: . $\ln(y) \;=\;\ln\bigg[823(1.044)x\bigg] \;=\;\ln(823) + \ln(1.044)^x \;=\;\ln(823) + x\!\cdot\!\ln(1.044)$

Then we have: . $\ln(y) \;=\;[\ln(1.044)]\!\cdot\! x + \ln(823)$

Let $m = \ln(1.044)$ ... a constant.
Let $b = \ln(823)$ ... another constant.

The equation becomes: . $\ln(y) \;=\; mx + b \quad\hdots\quad See?$