MHB Exponential and Logarithmic Functions

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The discussion revolves around solving the exponential function y = Ce^(kt) using given values to find constants k and C. Participants suggest using the provided data points, y = 100 when t = 2 and y = 300 when t = 4, to compute k. To find C, either of the two values can be substituted back into the equation. The conversation also hints at converting the exponential equation into logarithmic form for further analysis. The overall focus is on applying mathematical techniques to derive the constants from the exponential function.
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y=CektA) First find k. [Hint:Use the given information of y=100 when t=2, and y=300 when t=4 to compute k.]

B) Finally, find the value for C. [Hint use ine of the two pieces of information given in the problem to solve for C. in other words, use either y=100 when t=2 or use y=300 when 4=4 to compute C]. [Hint: to find t use y=1].
 
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qtheory said:
y=CektA) First find k. [Hint:Use the given information of y=100 when t=2, and y=300 when t=4 to compute k.]

B) Finally, find the value for C. [Hint use ine of the two pieces of information given in the problem to solve for C. in other words, use either y=100 when t=2 or use y=300 when 4=4 to compute C]. [Hint: to find t use y=1].
Is that a challenge problem? I'm heavily intoxicated at the moment so it's not obvious if it is or not.
 
qtheory said:
y=CektA) First find k. [Hint:Use the given information of y=100 when t=2, and y=300 when t=4 to compute k.]

B) Finally, find the value for C. [Hint use ine of the two pieces of information given in the problem to solve for C. in other words, use either y=100 when t=2 or use y=300 when 4=4 to compute C]. [Hint: to find t use y=1].

We are given:

$$y=Ce^{kt}$$

First, divide through by $C\ne0$ to get:

$$\frac{y}{C}=e^{kt}$$

Next convert from exponential to logarithmic form...what do you get?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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