(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Using the series expansion of e^KX and the fact that a^X=e^Xlna, evaluate 2^-3.4 accurate to 3 dp?

2. Relevant equations

3. The attempt at a solution

So a^X=e^Xlna. Basically we need to expand e^-3.4ln(2).

e^x=1+X+(X^2)/2!+(X^3)/3!...

e^-3.4ln(2)= 1-3.4ln(2)+(-3.4ln(2))^2/2+(-3.4ln(2))^3/6… See any errors in the expansion? The answer is supposedly 0.095 Can't seem to get it right.

Thank you.

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# Homework Help: Using the series expansion of e^KX, evaluate 2^-3.4 accurate to 3 dp?

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