Is there a test to verify that data is linear or rather logarithmic? I've done two small experiments, and if I visualize that data, it seems as there is a linear trend in the first one, and a logarithmic trend in the second. Is there a test to verify that this is indeed the case?
 A: There may be other ways to do it, but I would use a linear regression model.
We'll call your dependent variable in experiment 1 y1 and the independent variable x1. Using R code, I would first create a simple linear model:
M1 <- lm (y1 ~ x1)
summary(M1)

Hopefully, this shows a significant effect of x1 on y1. If not, if there is a linear trend, it is not significant. To show that the effect is linear and not non-linear, I could try a couple of other models and compare the models. The first one is quadratic (I use both x1 and x1^2), the second one is logarithmic (I use log(x1):
M2 <- lm (y1 ~ x1 + I(x1*x1))
M3 <- lm (y1 ~ x1 + log(x1)

I can now compare the models:
anova(M1, M2)
   anova(M1, M3)
And none of the tests should show a significant effect, meaning that the null hypohesis that M1 accounts for the data just as well as M2 or M3 cannot be rejected, and you can thus conclude that the relationship is linear.
For experiment 2, you do the same process but you try the logarithmic model first:
M4 <- lm (y2 ~ x2 + log(x2))
M5 <- lm (y2 ~ x2)
M6 <- lm (y2 ~ x2 + I(x2*x2))
anova (M4, M5)

And if you're correct, the first test should be significant, indicating that the model with more estimated parameters (M4) better accounts for the data. Now you cannot test M4 against M6 because the models are not nested (M4 and M5 are nested, because if the parameter estimate of log(x2) in M4 is set to 0, you have M5). But you can still compare them using the following command:
AIC(M4, M5, M6)

And M4 should be the lowest of them, and at least not much higher (say around 1 point or so) than M5 or M6.
You might want to use a model with log(x2) without including x2:
M7 <- lm(y2 ~ log(x2))

This is fine, but then M5 is no longer nested within M4. If M4 proves to be better than M5, then you can do:
anova(M4, M7)

These models are nested, and non-significant result will indicate that the null hypothesis that the simpler model (M7) best represents the data cannot be rejected. But if M4 isn't better than M5, you can do this:
AIC(M5, M6, M7)

And now, if M7 has the lowest AIC (by one point or so), you might conclude that a log-transformation of x provides a better fit to your data.
I don't know how easily these procedures can be done with other software packages than R.
I hope this helps.
