Question 718152
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Please help me solve for A and B in this exponential decay model y= ae^-bx. I have the following points: when x=0.6, y=1000 and when x=1.8, y=1.

I have done the following but I am not convienced it is the right path...

y=ae-bx therefor e-bx=y/a ... take log e of both sides ... loge e-bx = loge y/a

but logee=1 therefor -bx=loge y/a ... move the x over to get -b=x loge y/a

then enter values for x and y from example 1... -b= (0.6) loge(1000/a)

Is this the correct path to take for solving this?? Any advice would be greafull appreciated. Thank you.
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I've chosen to show you the CORRECT path, instead of looking through your work to identify your error(s), if any!

       {{{y = ae^(- bx)}}}, if: {{{system(matrix(2,1, x = 0.6, y = "1,000"))}}}           {{{y = ae^(-bx)}}}, if: {{{system(matrix(2,1, x = 1.8,  y = 1))}}}

          {{{system(matrix(3,1, "1,000" = ae^(- b(.6)), 
 "1,000" = ae^(- .6b),
"1,000"/e^(- .6b) = a))}}}               {{{system(matrix(3,1, 1 = ae^(- b(1.8)), 
1 = ae^(- 1.8b),
1/e^(- 1.8b) = a))}}}

Equating the 2 "a" values, we get: {{{"1,000"/e^(- .6b) = 1/e^(- 1.8b)}}}
                         {{{"1,000"(e^(- 1.8b)) = e^(- .6b)}}} ---- Cross-multiplying
                         {{{"1,000"(e^(- 1.8b))/e^(- 1.8b) = e^(- .6b)/e^(- 1.8b)}}} --- Dividing both sides by {{{e^(- 1.8b)}}}
                        {{{"1,000"cross((e^(- 1.8b)))/ cross(e^(- 1.8b)) = e^(- .6b)/e^(- 1.8b)}}} 
                                    {{{"1,000" = e^(- .6b - (- 1.8b))}}}
                                    {{{"1,000" = e^(- .6b + 1.8b)}}}
                                    {{{"1,000" = e^(1.2b)}}}
                                      1.2b = ln (1,000) --- Converting to LOGARITHMIC (natural) form 
                                        {{{highlight(b) = (ln ("1,000"))/1.2}}} = <font color = red><font size = 4><b>5.76, approximately.</font></font></b>

You can now substitute this value for b in either {{{a = "1,000"/e^(- .6b)}}}, or {{{a = 1/e^(- 1.8b)}}} to determine the value of "a."

If/When you do, you should get the value of a as approximately 31,622.8, to the nearest tenth.

@GREENESTAMPS is CORRECT! An INCORRECT value of 51,622.8 for "a" was entered previously, although the 
correct value, 31,622.8 was calculated. This has now been corrected. Thx for pointing out the error.</pre>