document.write( "Question 706998: the rate of interest for $2,000 invested for 10 years to accumulate to $3,000 if it is compounded monthly...using the formula a= p(1+r/n)^(nt) \n" ); document.write( "

Algebra.Com's Answer #435474 by josgarithmetic(39627) $\"\"$ $\"About$

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The information from the description is this:
\n" ); document.write( "p=2000
\n" ); document.write( "r=unknown
\n" ); document.write( "a=3000
\n" ); document.write( "n=12 (because the year is cut into 12 equal parts, meaning \"monthly\")
\n" ); document.write( "t=10\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Keep all as symbols, solve for r, AND THEN substitute the values.
\n" ); document.write( "As I OMIT most of the symbolic steps, the equation you will have is
\n" ); document.write( " $\"log%2810%2C%281%2Br%2Fn%29%29=%28log%2810%2Ca%29-log%2810%2Cp%29%29%2F%28nt%29\"$
\n" ); document.write( "(start by taking logarithm of both sides...).
\n" ); document.write( "... I am here referring to log base TEN (but you could choose whatever base you like).\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Do not be hung-up with the (1+r/n) number. It just represents the monthly interest rate. Just solve for (1+r/n), and when you have this value, you can then proceed to find r, the yearly rate, because you already know n. \n" ); document.write( "

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