document.write( "Question 1201094: You deposit $3000 into an account that earns 7% compounded annually. A friend deposits $2250 into an account that earns 6.85% annual interest, compounded continuously. Will your friend's balance ever equal yours? If so, when? If not, enter NEVER.\r
\n" ); document.write( "\n" ); document.write( "They will be equal in about
\n" ); document.write( " years (round to nearest whole year) \n" ); document.write( "
Algebra.Com's Answer #835346 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
your two investments are:
\n" ); document.write( "3000 * 1.07 ^ x
\n" ); document.write( "2250 * e^(.0685 * x)
\n" ); document.write( "they will be equal when 3000 * 1.07 ^x = 2250 * e^(.0685*x)
\n" ); document.write( "if you graph the difference, they will be equal when 3000 * 1.07^x - 2250 * e^(.0685 *x) = 0
\n" ); document.write( "the equivalent annual growth rate to e^(.0685*x) can be found as follows:
\n" ); document.write( "e^(.0685) = 1.070900625.
\n" ); document.write( "subtract 1 from that to get .070900625.
\n" ); document.write( "that's the equivalent annual growth rate.
\n" ); document.write( "you get 1.070900625 ^ x = e ^ (.0685 * x)
\n" ); document.write( "you can evaluate that for any x and you will see that the results will always be equal.
\n" ); document.write( "for example, when x = 40, you get 1.070900625^40 = 15.4869851 and e^(.0685*40) = the same.
\n" ); document.write( "you can do that for any x and the resulting values will be the same.
\n" ); document.write( "since e^(.0685*x) is equivalent to 1.070900625^x, then you can substitute 1.070900625^x for e^(.0685*x)
\n" ); document.write( "since the euivalent annual growth rate is higher than 1.07, it suggests that 3000 * 1.07^x will be equal to 2250 * 1.070900625^x at some point in time.
\n" ); document.write( "start with 3000 * 1.07^x = 2250 * 1.070900625^x
\n" ); document.write( "divide both sides of that equation by 2250 and divide both sides of that equation by 1.07^x to get:
\n" ); document.write( "3000/2250 = 1.070900625^x/1.07^x
\n" ); document.write( "take the log of both sides of that eqution to get:
\n" ); document.write( "log(3000/2250) = log(1.070900625^x/1.07^x)
\n" ); document.write( "by log rule that says log(a/b) = log(a) - log(b), that becomes:
\n" ); document.write( "log(3000/2250) = log(1.070900625^x) - log(1.07^x)
\n" ); document.write( "by log rule that says log(b^x) = x * log(b), that becomes:
\n" ); document.write( "log(3000/2250) = x * log(1.070900625) - x * (log(1.07)
\n" ); document.write( "factor out the x to get:
\n" ); document.write( "log(3000/2250) = x * (log(1.070900625) - log(1.07)
\n" ); document.write( "by log rule that says log(a) - log(b) = log(a/b), that becomes:
\n" ); document.write( "log(3000/2250) = x * log(1.070900625/1.07)
\n" ); document.write( "divide both sides of the equation by log(1.070900625/1.07) to get:
\n" ); document.write( "log(3000/2250) / log(1.070900625/1.07) = x
\n" ); document.write( "solve for x to get:
\n" ); document.write( "x = 341.9285085
\n" ); document.write( "3000 * 1.07^x will be equal to 2025 * 1.070900625 ^ x when x = 341.9285085.
\n" ); document.write( "since 1.070900625 ^ x is equivalent to e^(.0685*x), this also says that:
\n" ); document.write( "3000 * 1.07^x will be equal to 2025 * e^(.0685*x) when x = 341.9285085.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "3000 * 1.07 ^ 341.9285085 = 3.44048223 * 10^13.
\n" ); document.write( "2250 * 1.070900625 ^ 341.9285085 = 3.44048223 * 10^13.
\n" ); document.write( "2250 * e^(.0685 * 341.9285085) = 3.44048223 * 10^13.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "all 3 equations are equal when x = 341.9285085.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "your solution is that 3000 * 1.07 ^ x will be equal to 2250 * e^(.0685 * x) in 341.9285085 years.
\n" ); document.write( "round to nearest whole year to get 342 years.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "graph y = 3000 * 1.07^x - 2250 * e^(.0685*x) and graph y = 3000 * 1.07^x - 2250 * 1.070900625 ^ x, as shown below.
\n" ); document.write( "you will see that the graph crosses y = 0 when x = 341.9285085 for both equations.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "here's the graph.
\n" ); document.write( "\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );