I have figure which is logarithmic scale on both axis. There's a line on that figure, I know two points on that line and want to interpolate a third point on that line based on the two known points.

(What you see basically is a curve which is constituted of linear segments. I'll only do interpolation within such a linear segment, knowing the two boundary points of the segment. The input value (flow) is on the $y$ axis, and I'm looking for the $x$ value related to that.

There's a linear interpolation (by computing fractions), but my recent try on logarithmic interpolation sometimes produces worse results than the linear one. Disappointing.

First I tried to follow which was a disaster.

Then I tried to think (My Take: on the Google Drawing). What am I missing? The outcome is close to the curve, but I want an interpolation which fits a point spot on to the line segment. Mathematically that should be possible.

I feel like I'm missing an logarithmic/exponential part somewhere, but I don't know where.

My Take if the two end point of the line segment is $(x_1, y_1)$ and $(x_2, y_2)$, the measurement $y_3$, and I want to know $x_3$ (see Google Drawing): \begin{align} y_1 &= 10^{ax_1 + b}\\ y_2 &= 10^{ax_2 + b}\\ \log(y_1) &= ax_1 + b \\ \to b &= \log(y_1) - ax_1\\ \log(y_2) &= ax_2 + b \\ \to a &= \frac{\log(y_2) - b}{x_2}\\ &= \frac{\log(y_2) - (\log(y_1) - ax_1)}{x_2}\\ a &= \frac{\log(y_2) - \log(y_1) + ax_1}{x_2}\\ ax_2 - ax_1 &= \log(y_2) - \log(y_1)\\ a &= \frac{\log(y_2) - \log(y_1)}{x_2 - x_1}\\ y_3 &= 10^{ax_3 + b}\\ x_3 &= \frac{\log(y_3) - b}{a} \end{align}

@John pointed to the right direction and the Wiki article helped. The difference is that I'm looking for $x_3$. So the

$$y_3 = y_1 * \left(\frac{x_3 }{ x_1}\right) ^ m$$

equation transforms to:

$$x_3 = x_1 * \left(\frac{y_3 }{ y_1}\right) ^ {1/m}$$