Question 547070
At the end of 1990:
{{{ P = 226000 }}}
At the end of 1991:
{{{ P = 226000 + .025*226000 }}}
{{{ P = 226000*( 1 + .025 ) }}}
At the end of 1992:
{{{ P = 226000*( 1 + .025 ) + .025*226000*( 1 + .025 ) }}}
{{{ P = ( 1 + .025 )*226000*( 1 + .025 ) }}}
{{{ P = 226000*( 1 + .025 )^2 }}}
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I can say:
{{{ 1000000 = 226000*( 1 + .025 )^n }}}
Where {{{ n=0 }}} is 1990
{{{ n = 1 }}} = 1991, etc.
{{{ ( 1 + .025 )^n = 1000000/226000 }}}
{{{  1.025^n = 4.425 }}}
{{{ n*log( 1.025 ) = log( 4.425 ) }}}
{{{ n = log( 4.425 ) / log( 1.025 ) }}}
You can find with calculator
Add n to 1990
The decimal portion gets multiplied by 12 to find month