Question 207168
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If interest is continously compounded, then we use the formula,
{{{F=Pe^(rt)}}}
where{{{system(F=Future=1200, P=present=600, r=rate=0.0564, t=time_in_years)}}}


Then,
{{{1200=600e^(0.0564*t)}}} ---> {{{cross(1200)2/cross(600)=cross(600)e^(0.0564*t)/cross(600)}}}
{{{2=e^(0.0564*t)}}}
{{{ln2=0.0564*t}}} ---> {{{ln2/0.0564=cross(0.0564)t/cross(0.0564)}}}
{{{highlight(t=12.29yrs)}}}, Answer  or <font color=blue> 12 yrs, 3 mos. & 14 days</font>


Thank you,
Jojo</font>