Question 961552
P=pe^(kt), using small p to avoid using subscripts.  Big P is for balance of the bond after any time t.   The formula rendered is {{{P=pe^(kt)}}}.  Your goal is in part, to isolate p, which you can do multiplying both sides by {{{e^(-kt)}}}.  


If p=1, the in 1 year, t=1, {{{P=1*e^(0.03*1)}}}, {{{P=e^(0.03)=1.0304}}}, reasonably near the expected 3% increase.  The value of constant k is the same as the interest rate of 3% or 0.03=r=k.


Your description of P=10000 after t=10 means you still want to know the initial investment amount.
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ln(P)=ln(p)+ln(e^(kt))
ln(P)=ln(p)+kt*1
ln(P)-kt=ln(p)
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those steps seem unnecessary although correct.



BETTER-----------------------------------------
{{{pe^(kt)=P}}}
{{{p=P/e^(kt)}}}-----you could use a calculator for this.
{{{p=10000/(e^(0.03*10))}}}
{{{p=10000/(e^(0.3))}}}
{{{highlight(p=7408.18)}}}, the initial investment amount.


Just 5 years instead of 10 years?
{{{P=7408.18*e^(0.03*5)}}}
{{{P=7408.18*1.1618342427}}}
{{{highlight(P=8607.08)}}}