Question 697434
For now, let {{{ p }}} = the purchase price
The value is:
1st yr: {{{ p }}}
2nd yr: {{{ .7p }}}
3rd yr: {{{ .7*( .7p ) }}}
4th yr: {{{ .7^3*p }}}
etc
Let {{{ V }}} = the current value of the laptop
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If {{{ n }}} the number of years,
{{{ V = .7^(n - 1)*p }}}
( I use the {{{ n-1 }}} because it makes the 
1st year {{{ .7^( 1-1)*p = p }}} )
{{{ p = 1200 }}}
{{{ V = 300 }}}
{{{ 300 = .7^( n-1 )*1200 }}}
{{{ .7^( n-1 ) = 1/4 }}}
{{{ .7^( n-1 ) = .25 }}}
Take the log base 10 of both sides
{{{ log( .7^( n-1 ) ) = log( .25 ) }}}
{{{ ( n-1 )*log( .7 ) = log( .25 ) }}}
{{{ n-1 = log( .25 ) / log( .7 ) }}}
{{{ n =  log( .25 ) / log( .7 ) + 1 }}}
{{{ n = ( -.60206 ) / ( -.1549 ) + 1 }}}
{{{ n = 3.8868 + 1 }}}
{{{ n = 4.8868 }}}
After the 4th year, the laptop will be
worth $300 or less.
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Check answer:
{{{ V = .7^( n-1 )*1200 }}}
{{{ V = .7^(4.8868-1)*1200 }}}
{{{ V = .7^3.8868*1200 }}}
{{{ V = .24999*1200 }}}
{{{ V = 300 }}}
OK