Answer:
Therefore 6th year his income was $19,700.
Step-by-step explanation:
Given, Joe's annual income has been increasing in some dollar amount . The first year his income was $15,200 and 4th year his income was $17,900.
A=$17900, P= $15,200 and n = 3 year
[tex]A= P(1+\frac{r}{100} )^n[/tex]
[tex]\Leftrightarrow 17900=15200(1+\frac{r}{100})^3[/tex]
[tex]\Leftrightarrow (1+\frac{r}{100})^3=\frac{17900}{15200}[/tex]
[tex]\Leftrightarrow (1+\frac{r}{100})=(\frac{17900}{15200})^{\frac{1}{3} }[/tex]
[tex]\Leftrightarrow \frac{r}{100}=(\frac{17900}{15200})^{\frac{1}{3} }-1[/tex]
[tex]\Leftrightarrow r=5.60[/tex]
Let [tex]t^{th}[/tex] year Joe's income was $19,700.
[tex]\therefore 19700=15,200(1+\frac{5.60}{100} )^{t-1}[/tex]
[tex]\Leftrightarrow 1.296= (1+0.056)^{t-1}[/tex]
[tex]\Leftrightarrow 1.296= (1.056)^{t-1}[/tex]
[tex]\Leftrightarrow log(1.296)= (t-1)log(1.056)[/tex]
[tex]\Leftrightarrow \frac{log(1.296)}{log(1.056)}= (t-1)[/tex]
[tex]\Leftrightarrow t-1= 4.75[/tex]
[tex]\Leftrightarrow t= 4.75+1[/tex]
[tex]\Leftrightarrow t = 5.75[/tex] ≈6
Therefore 6th year his income was $19,700.
