Answer
$y=2.5e^{-0.357x}$
Work Step by Step
Use Basic Logarithmic Properties: $b^{\log_{b}x}=x$,
replacing b with e and x with $(0.7)^{x}$
$y=2.5(0.7)^{x}$ is equivalent to $y=100e^{\ln(0.7)^{x}}$\qquad
... now, by the Power Rule: $\log_{\mathrm{b}}\mathrm{M}^{\mathrm{p}}=\mathrm{p}\log_{\mathrm{b}}\mathrm{M}$
$y=2.5e^{(\ln 0.7)x}$;
Calculator:
$\ln 7.3\approx$-0.356674943939 $\qquad$ ... round to 3 decimals
$\approx-0.357$,
$y=2.5e^{-0.357x}$.
