Question 23218
WHAT A NEAT PROBLEM!!  NICE!!


For Canada {{{y=30 e^(.014t) }}} in millions
For Germany {{{y = 80 e^(-.017t) }}} in millions


When will populations be equal?
{{{30 e^(.014t) =  80 e^(-.017t) }}} 


Divide both sides by  {{{30 e^(.014t)}}}:
{{{(30 e^(.014t))/(30e^(.014t))  =  (80 e^(-.017t))/(30e^(.014t)) }}}
{{{ 1  =  (80 e^(-.017t))/( 30e^(.014t)) }}}


When you divide with the same base number, you subtract exponents:
{{{1  = (8/3)  e^(-.017t- .014t) = (8/3)*e^(-.031t) }}}


Multiply both sides by {{{3/8}}}
{{{3/8 = e^(-.031t)}}}

Take the ln of each side:
{{{ln(3/8) = ln(e^(-.031t)) }}}
{{{ln (3/8) = -.031t}}}

{{{ t= (ln(3/8))/ -.031}}}
{{{t= 31.639655}}} calculator accuracy

Rounds off to 31.64 years.


R^2 at SCC