Can logarithm bases 2 and 10 both be expressed as $e$ in $\log$ / $\ln$ entropy?

Question

If

$$\exp\left({\sum_x p(x)\log_e \frac{1}{p(x)}}\right) = 2^{-\sum_x p(x)\log_2 p(x)}$$

then does

$$\exp\left({\sum_x p(x)\ln \frac{1}{p(x)}}\right) = 10^{-\sum_x p(x)\ln p(x)}?$$

Finally, can both of the above RHS's be expressed as the formula below, with an e at the front instead of $10$ or $2$, or only one? if not both, which one?

$$e^{-\sum_x p(x)\log_b p(x)}$$

In other words, does $$e^{-\sum_x p(x)\log p(x)} = 2^{-\sum_x p(x)\log p(x)}$$

and

$$e^{-\sum_x p(x)\ln p(x)} = 10^{-\sum_x p(x)\ln p(x)}?$$

Answer 1

For any base $b$, $$ b^{-\sum_x p(x) \log_b p(x)} = \prod_x b^{-p(x) \log_b p(x)} = \prod_x (b^{\log_b p(x)})^{-p(x)} = \prod_x p(x)^{-p(x)} $$ which does not depend on $b$. So all these expressions are equal.

However, the base of the exponent must match the base of the log.