document.write( "Question 6356: A=$6000 P= $1000 interest is 12% compounded quarterly Find # of interest periods \n" ); document.write( "
Algebra.Com's Answer #3407 by AnlytcPhil(1810) You can put this solution on YOUR website! A=$6000 P= $1000 interest is 12% compounded quarterly \n" ); document.write( "Find # of interest periods \n" ); document.write( "` \n" ); document.write( "The formula is \n" ); document.write( "` \n" ); document.write( "A = P(1+r/n)nt \n" ); document.write( "` \n" ); document.write( "Where \n" ); document.write( "` \n" ); document.write( "A = ending amount = 6000 \n" ); document.write( "P = beginning amount = 1000 \n" ); document.write( "r = annual rate expressed as a decimal = .12 \n" ); document.write( "n = number of interest periods per year = 4 \n" ); document.write( "` ` (since it's quarterly) \n" ); document.write( "t = number of years (which we find first) \n" ); document.write( "` \n" ); document.write( "A = P(1+r/n)nt \n" ); document.write( "` \n" ); document.write( "Taking logs of both sides \n" ); document.write( "` \n" ); document.write( "logA = log[P(1+r/n)nt] \n" ); document.write( "` \n" ); document.write( "logA = logP + log(1+r/n)nt \n" ); document.write( "` \n" ); document.write( "logA = logP + nt[log(1+r/n)] \n" ); document.write( "` \n" ); document.write( "logP + nt[log(1+r/n)] = logA \n" ); document.write( "` \n" ); document.write( "nt[log(1+r/n)] = logA - logP \n" ); document.write( "` \n" ); document.write( "nt[log(1+r/n)] `` logA - logP \n" ); document.write( "-------------- = ------------- \n" ); document.write( "`n[log(1+r/n)] ` n[log(1+r/n)] \n" ); document.write( "` \n" ); document.write( "nt[log(1+r/n)] `` logA - logP \n" ); document.write( "-------------- = ------------- \n" ); document.write( "`n[log(1+r/n)] ` n[log(1+r/n)] \n" ); document.write( "` \n" ); document.write( "` ` ` ` ` ` ` ``` logA - logP \n" ); document.write( "` ` ` ` ` ` `t = ------------- \n" ); document.write( "` ` ` ` ` ` ` `` n[log(1+r/n)] \n" ); document.write( "` \n" ); document.write( "` ` ` ` ` ` ` ``` log6000 - log1000 \n" ); document.write( "` ` ` ` ` ` `t = -------------------- \n" ); document.write( "` ` ` ` ` ` ` ` ` `4[log(1+.12/4)] \n" ); document.write( "` \n" ); document.write( "` ` ` ` ` ` ` ``` 3.77815125 - 3 \n" ); document.write( "` ` ` ` ` ` `t = -------------------- \n" ); document.write( "` ` ` ` ` ` ` ` ` `4[log(1.03)] \n" ); document.write( "` \n" ); document.write( "` ` ` ` ` ` ` `` ` .7781512504 \n" ); document.write( "` ` ` ` ` ` `t = --------------- \n" ); document.write( "` ` ` ` ` ` ` ` ` `.0513488988 \n" ); document.write( "` \n" ); document.write( "` ` ` ` ` ` ` `` ` .77815125 \n" ); document.write( "` ` ` ` ` ` `t = --------------- \n" ); document.write( "` ` ` ` ` ` ` ` ` `.0513488988 \n" ); document.write( "` \n" ); document.write( "` ` ` ` ` ` `t = 15.15419548 years \n" ); document.write( "` \n" ); document.write( "Multiply by 4 to get the number of interest periods \n" ); document.write( "` \n" ); document.write( "Answer = 60.61678192 which means that 60 interest periods will \n" ); document.write( "yield slightly less that $6000, or $5891.60 and 61 interest period \n" ); document.write( "will yield slightly more than $6000, or $6068.40. \n" ); document.write( "` \n" ); document.write( "Edwin J\r \n" ); document.write( "\n" ); document.write( " \r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |