Simplifying $\log A\cdot\log B$

Question

We know that $\log A\cdot B = \log A+\log B$. By virtue of the change of base rule, we know that $\log_BA=\frac{\log A}{\log B}$. But is there any way we can further simplify, or rewrite products of two logarithms? $$\log A\cdot\log B=?$$

I suppose it can be written as $$\log A\cdot\log B=\log A^{\log B}$$ But there isn't anything special about it.

But should there be any at all?

Answer 1

Regarding the existence of other properties, the property:

$$\log_b\left(A^B\right) = B\cdot\log_b(A)$$

Comes from the fact that:

$$\left(b^{\,A}\right)^B=\left(b^{\,A\cdot B}\right)$$

In other words, the multiplicative power rule. There really isn't another way to further simplify that, the log property just stems from a property of exponents.

Answer 2

$$\log A\cdot\log B=\log A^{\log B}$$ $$\log A\cdot\log B=\log B^{\log A}$$ $$\log A^{\log B}=\log A\cdot\log B=\log B^{\log A}$$ $$\log A^{\log B}=\log B^{\log A}$$ New identity $$A^{\log B}=B^{\log A}$$ $$A^{\ln B}=B^{\ln A}$$ $$A^{\log_CB}=B^{\log_CA}$$