How to get $d$ in terms of $A$ and $B$

Question

I'm trying to get $d$ in terms of $A$ and $B$ having the next equations:

$$0 = A + B*\log _2(d)$$ $$6 = A + B*\log _2(\frac{d}{2})$$

EDIT

How about $A$ in terms of $B$ and $d$? And $B$ in terms of $A$ and $d$?

score 1 · Answer 1 · answered Oct 06 '15 at 16:22

1

Hint.

$$ \log_2 \left( \frac{d}{2} \right) = \log_2 d - \log_2 2 = \log_2 d - 1 $$

Incidentally, a common abbreviation for $\log_2 x$ is $\lg x$.

answered Oct 06 '15 at 16:22

Brian Tung

Thanks! How about $A$ in terms of $B$ and $d$? And $B$ in terms of $A$ and $d$? – user2952272 Oct 06 '15 at 18:59
If you have $A$ and $B$ in terms of $d$, can't you use simple algebra to obtain the other things you want? – Brian Tung Oct 06 '15 at 19:53

score 1 · Answer 2 · answered Oct 06 '15 at 16:23

1

The first equation gives you directly

$$d=2^{-\frac{A}{B}}$$

answered Oct 06 '15 at 16:23

Tryss

Thanks! How about $A$ in terms of $B$ and $d$? And $B$ in terms of $A$ and $d$? – user2952272 Oct 06 '15 at 18:59

2 Answers2