Simplifying $\exp(a)\ln(b)$

Question

I am working on simplifying a problem, and I ran into $\exp(a)\ln(b)$. Here's what I tried:
$$\exp(a)\ln(b)$$ $$\ln(\exp(\exp(a)))\ln(b)$$ I also tried:
$$\exp(a)\ln(b)$$ $$\exp(a)\exp(\ln(\ln(b)))$$ $$\exp(a+\ln(\ln(b)))$$ Are there any other formulas for $\exp(a)\ln(b)$?

Answer 1

I don't believe your first expression is equivalent, for example:

$$e^2 \cdot \ln(3) = 8.12$$

$$\ln(e^{e^2} + 3) = 7.39$$

This is because $(\ln x)(\ln y) \neq \ln(x+y),$ which you did in your second step. As for finding a more simple expression, I think the one you started with is pretty much as good as it gets. Some other options I've devised for you include:

$$e^a \ln b = \ln(e^{e^a}) \ln b = \ln\left((e^{e^a})^{\ln b}\right) = \ln \left(e^{(e^{e^a})^{\ln b}}\right)$$

which, after enough reducing, gives me $\ln b \ \cdot \ e^a$

Answer 2

The simplest form is $e^{a}\ln(b)$.

Following from what you've done, we have: $e^{a}\ln(b)$=$e^{a+\ln(\ln(b))}$=$\ln(e^{e^{a}})\ln(b)$=$e^{\ln(\ln(b)e^{a})}$=$\ln(e^{\ln(b)e^{a}})$.