Question 945779
{{{ A }}} is the amount you end up with
after time {{{ t }}}
{{{ A = 1000 + 100 = 1100 }}}
{{{ P }}} = amount you started out with
{{{ P = 1000 }}}
{{{ n = 4 }}} because of compounding
quarterly
{{{ r = .04 }}} is a 4% interest rate
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{{{ 1100 = 1000*( 1 + .04/4 )^(4t) }}}
{{{ 1100 = 1000*( 1.01)^(4t) }}}
Divide both sides by {{{ 1000 }}}
{{{ 1.1 = ( 1.01 )^(4t) }}}
Take the log base {{{ 10 }}} of both sides
{{{ log(( 1.1 )) = log(( 1.01^(4t) )) }}}
{{{ log(( 1.1 )) = 4t*log(( 1.01 )) }}}
{{{ 4t = log(( 1.1)) / log(( 1.01 )) }}}
{{{ t = log(( 1.1 )) / ( 4*log(( 1.01 ) )) }}}
{{{ t = .041393 / ( 4*.00432137 ) }}}
{{{ t = .041393 / .0172855 }}}
{{{ t = 2.394666 }}}
2.394666 years
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check:
{{{ 1100 = 1000*( 1 + .04/4 )^(4t) }}}
{{{ 1100 = 1000*( 1 + .04/4 )^(4*2.394666) }}}
{{{ 1100 = 1000*( 1 + .04/4 )^(9.578664) }}}
{{{ 1.1 = 1.01^9.578664 }}}
{{{ 1.1 = 1.1000008 }}}
close enough