# How to get d in terms of A and B 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(d/2) EDIT How about A in terms of B and d? And B in terms of A and d?

How to get $d$ in terms of $A$ and $B$
I'm trying to get $d$ in terms of $A$ and $B$ having the next equations:
$0=A+B\ast {\mathrm{log}}_{2}\left(d\right)$
$6=A+B\ast {\mathrm{log}}_{2}\left(\frac{d}{2}\right)$
EDIT
How about A in terms of B and d? And B in terms of A and d?
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indyterpep
Hint.
${\mathrm{log}}_{2}\left(\frac{d}{2}\right)={\mathrm{log}}_{2}d-{\mathrm{log}}_{2}2={\mathrm{log}}_{2}d-1$
Incidentally, a common abbreviation for ${\mathrm{log}}_{2}x$ is $\mathrm{lg}x$
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Kevin Charles
The first equation gives you directly
$d={2}^{-\frac{A}{B}}$