0
$\begingroup$

I am trying to solve for $x$ in this equation: $$ab-c+2^x=2^xd$$ I've been trying to use Maple to figure it out but to no avail.

Looking for the mathematical solution and if possible a way to get the step-by-step solution in Maple also.

enter image description here

$\endgroup$
6
  • 2
    $\begingroup$ Subtract $2^x$ from both sides, and divide both sides by $d-1$? Then take a logarithm? $\endgroup$
    – angryavian
    Oct 13 at 21:40
  • $\begingroup$ You might need to solve a variable first, as thats what it says on there. $\endgroup$ Oct 13 at 21:41
  • $\begingroup$ @angryavian, thanks, sorry may not have been clear but I need to solve for x, if i subtract 2^x from both sides what happens to the x then? $\endgroup$ Oct 13 at 21:46
  • 1
    $\begingroup$ On the left, the $2^x$ goes away. On the right, since you have $d$ copies of $2^x$ and you are subtracting one, you now have $d - 1$ copies of it. So, on the right you have $2^x(d - 1)$. If you divide both sides by $(d - 1)$, that leaves $2^x$ on the right-hand side. If you take the base 2 logarithm, that will leave just $x$. $\endgroup$
    – johnnyb
    Oct 13 at 21:51
  • $\begingroup$ @johnny, thanks a lot $\endgroup$ Oct 13 at 22:03
0
$\begingroup$

$$\begin{align} &ab-c+2^x=2^xd\\ &\Rightarrow{ab-c}=2^{x}\left({d-1}\right)\\ &\Rightarrow{2^{x}}=\left(\frac{ab-c}{d-1}\right)\\&\Rightarrow{x}=\log_2\left(\frac{ab-c}{d-1}\right)\end{align}$$

The main conditions are: $$ {d}\neq{1},\quad\left(\frac{ab-c}{d-1}\right)\gt{0}. $$

Good luck!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.