Choose the regression direction for an inverse prediction

Plot all six pairs with dosage $x$ horizontally and pulse-rate decrease $y$ vertically. The increasing pattern bends downwards; see the comparison sketch.

$r_{x,y}=0.9691927765\approx0.969$. A value close to 1 does not exclude curvature: the scatter indicates an increasing relationship with a decreasing gradient.

$y=\ln(cx^d)=\ln c+d\ln x$. The correlation of $\ln x$ and $y$ is $0.9923885447\approx0.992$, closer to 1 than the raw value. This and the decreasing-gradient pattern support the transformed model.

Regress $y$ on $u=\ln x$ to obtain $y=4.841951669+11.98564439u$. Therefore $\ln c=4.841951669$, so $c=e^{4.841951669}\approx127$; $d=11.98564439\approx12.0$ (3 s.f.).

To estimate $x$ from a given $y$ by regression, fit $u=\ln x$ on $y$, not $y$ on $u$. The line is $u=-0.3833133126+0.08216788280y$. At $y=25$, $\hat x=e^{-0.3833133126+0.08216788280(25)}=5.31686454\approx5.32\ \mathrm{cm}^3$. The printed solution instead solves $25=4.84195+11.98564\ln x$, giving $x=5.38\ \mathrm{cm}^3$. That is inversion of the forward fitted model, not the reverse least-squares regression; the two predictions differ because $|r|<1$. Both extrapolate beyond the observed decrease range $[9,21]$ and are unreliable.