Q:

Kloh put $3000 dollars in a savings account that earns 4% annually, compounded monthly. use logarithms to find how long would it would take for her to double her money?​

Accepted Solution

A:
Answer:[tex]17.4\ years[/tex]  Step-by-step explanation:we know that    The compound interest formula is equal to  [tex]A=P(1+\frac{r}{n})^{nt}[/tex]  where  A is the Final Investment Value  P is the Principal amount of money to be invested  r is the rate of interest  in decimal t is Number of Time Periods  n is the number of times interest is compounded per year in this problem we have  [tex]t=?\ years\\ P=\$3,000\\ r=0.04\\n=12\\ A=\$6,000[/tex]  substitute in the formula above  [tex]\$6,000=\$3,000(1+\frac{0.04}{12})^{12t}[/tex]  [tex]2=(\frac{12.04}{12})^{12t}[/tex]  Applying log both sides[tex]log(2)=log[(\frac{12.04}{12})^{12t}][/tex]  [tex]log(2)=(12t)log[(\frac{12.04}{12})][/tex]  [tex]t=log(2)/[(12)log(\frac{12.04}{12})]=17.4\ years[/tex]