Where am i wrong in this calculation of $A\log_2a+B\log_2b = (\log_2 ab^B)$

Question

\begin{align} A\log_2a+B\log_2b & = A(\log_2a+\frac{B}{A}\log_2b) \\ & = A(\log_2a+\log_2b^{\frac{B}{A}})\\ & =A(\log_2 ab^{\frac{B}{A}}) \\ & = (\log_2 ab^{\frac{B}{A}})^A \\ & = (\log_2 ab^B) \end{align}

I feel there is something strange,because the sum of the left hand side of formula has a relation with A, I mean if A become smaller or bigger, the sum of left hand side will become smaller or bigger too.

However, it seems that the sum of right hand side, $\log_2ab^B$, doesn't have any relation with A. So I think I make a mistake in somewhere, can anyone tell me where am I wrong?

Answer 1

Last two lines are both wrong. $A(\log_2(a b^{\frac B A}))=\log_2 ((a b^{\frac B A})^{A})=\log_2(a^{A}b^{B})$

Answer 2

You really have:

$\begin{align*} A \log_2 a + B \log_2 b &= \log_2 a^A + \log_2 b^B \\ &= \log_2 (a^A b^B) \end{align*}$

Answer 3

The error is that $ab^{B/A}$ means that $b$ is raised to the power of $B/A$, but in the last step you seem to interpret it as $(ab)^{B/A},$ which is not the same. That is, in general $(ab^{B/A})^A \ne ab^B$

Another error is is the penultimate step where you should have written $\log_2 ((ab^{B/A})^A)$, but that's minor since you seem to interpret it as that.

Answer 4

$$ A \log a + B \log b = \log \big(a^A\big) + \log \big(b^B\big) = \log \big(a^A \cdot b^B\big) $$