How do I convert (ln) y= a*b^x to base of e??eg; ln y = 3.93 x 0.99^x?

Question

am studying newtons law of cooling, need to find one equation 2 different ways, have found it using regression function and got y = 49.65e^(-0.027x) and found ln y = 3.93 x 0.99^x using the exponential function now how do i make this equation match the first one in order to prove it is correct????

Answer 1

This was deviously tricky so I had to fight fire with fire.

First I graphed, then ran a Newton iteration on the equation :

y = ln[49.65e^(-0.027x)] - 3.93*(0.99)^x

and found what seems to be the only root at x = 2.0679...

I'm presuming that x near 2 may mean something to you ??

Or it may not!

For numbers not too far removed from x = 2,

the following trick works quite well, to 2 decimal places..

0.99^0 = 1.00

0.99^1 = 0.99

0.99^2 = 0.98

0.99^3 = 0.97

0.99^4 = 0.96

0.99^5 = 0.95

0.99^6 = 0.94

0.99^7 = 0.93

0.99^8 = 0.92

0.99^9 = 0.91

0.99^10 = 0.90

So, 0.99^x = 1 - 0.01x,

or close enough, at least for 0 <= x <= 10.

This comes from the fact that 0.99^x = (1 - 0.01)^x,

which when expanded, shows that only the first two

terms are the most significant, due to the fact that

0.01 is only 1/100 of 1, so subsequent terms only

play a minor part until x becomes larger.

Now, starting with :

ln(y) = 3.93 * 0.99^x

Substituting 1 - 0.01x for 0.99^x gives :

ln(y) = 3.93 * (1 - 0.01x)

Multiplying out :

ln(y) = 3.93 - 0.0393x

Now, 3.93 = ln(50.91), and -0.0393x = ln[e^(-0.0393x)]

So, ln(y) = ln(50.91) + ln[e^(-0.0393x)]

Therefore, ln(y) = ln[50.91 * e^(-0.0393x)]

Thus, y = 50.91e^(-0.0393x)

which is similar to y = 49.65e^(-0.027x).

Plotting the graphs shows their similar shape.

They intersect at about x = 2, then diverge for

quite some way before finally converging again.

Their greatest difference is a little less than 6.5

units on the Y-axis, when x is about 33.

Up to x = 10, the greatest difference is about 3.5.