...and if *e* is the constant that is Euler's number, then *a*'s approximate value can be found to umpteen decimal places.

Also, y=ae^(-k*x) is different from the OP's y=ae^-k*x: the two forms don't give the same answer...as shown by MIF's calculator. Btw, the calculator treats 'e' as being Euler's number.

]]>Bob

]]>]]>Can anyone assist with the following problem to find a where x = 0.85, k=1 and y = 11.2;

y=ae^-k*x

So 11.2 = a^(-0.85)

Welcome to the forum.

Can anyone assist with the following problem to find a where x = 0.85, k=1 and y = 11.2;

y=ae^-k*x

Please see next post by Alg Num Theory. I'll leave mine as a testament to the importance of 'reading the question' !!!

So 11.2 = a^(-0.85)

If you take logs (any will do but let's use base 10 as it's handy on a calculator) you get

log(11.2) = log(a^-0.85) = -0.85 * log(a)

So re-arrange this to make log(a) the subject, and inverse log it (ie. raise 10 to the power of) and you'll get a.

eg. If log(a) = 2 then a = 10^2 = 100

Bob

]]>y=ae^-k*x

Thank you

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