Question

Joe's annual income has been increasing in the same dollar amount. The first year his income was $15,200, and the 4th year his income was $17,900. In which year was his income $19,700?​

Answer 1

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

$A= P(1+\frac{r}{100} )^n$

$\Leftrightarrow 17900=15200(1+\frac{r}{100})^3$

$\Leftrightarrow (1+\frac{r}{100})^3=\frac{17900}{15200}$

$\Leftrightarrow (1+\frac{r}{100})=(\frac{17900}{15200})^{\frac{1}{3} }$

$\Leftrightarrow \frac{r}{100}=(\frac{17900}{15200})^{\frac{1}{3} }-1$

$\Leftrightarrow r=5.60$

Let  $t^{th}$ year Joe's income was $19,700.

$\therefore 19700=15,200(1+\frac{5.60}{100} )^{t-1}$

$\Leftrightarrow 1.296= (1+0.056)^{t-1}$

$\Leftrightarrow 1.296= (1.056)^{t-1}$

$\Leftrightarrow log(1.296)= (t-1)log(1.056)$

$\Leftrightarrow \frac{log(1.296)}{log(1.056)}= (t-1)$

$\Leftrightarrow t-1= 4.75$

$\Leftrightarrow t= 4.75+1$

$\Leftrightarrow t = 5.75$ ≈6

Therefore 6th year his income was $19,700.