DC CAPE Algebra II Linear, Quadratic, and Exponential Models. Practice it free.

$10^{3} = 1{,}000\textsf{,}$ so $\log_{10}1{,}000 = 3\textsf{.}$

The remaining amount after $t$ years is $A(t) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{t}{h}}\textsf{,}$ where $h$ is the half-life. Here $t = 10$ and $h = 5\textsf{,}$ so $A(10) = 80\left(\dfrac{1}{2}\right)^{\textstyle\frac{10}{5}} = 80\left(\dfrac{1}{2}\right)^{2} = 20\textsf{.}$

When $t = 1$, $I(1) = 275(1.04)^1 = 275 \cdot 1.04 = 286$ thousand dollars.

$400 \times 1.25 = 500$.

Slope-intercept form is $f(x) = mx + b\textsf{.}$ The height decreases by $2$ centimeters each hour, so $m = -2\textsf{.}$

The outputs increase by $3$ each time $x$ increases by $1\textsf{,}$ so the constant rate is $3\textsf{.}$ Using the point $(1, 7)$ gives $3(1) + b = 7\textsf{,}$ so $b = 4\textsf{.}$ Therefore $h(x) = 3x + 4\textsf{.}$

Point-slope form is $y - y_1 = m(x - x_1)$. With $m = 4$ and $(x_1, y_1) = (-1, 2)$, the equation is $y - 2 = 4(x - (-1))$, or $y - 2 = 4(x + 1)\textsf{.}$

Take $\ln$ of both sides: $-x=\ln 6$. Since $-x=(-1)x$, divide both sides by $-1$ to get $x=-\ln 6\textsf{.}$