Question

Bet u can’t solve this

Answer 1

Answer:

Step-by-step explanation:

Assuming you're solving for p:

$m=2^{-1} *2^p(2^p-1)$

Let $y=2^p$

Now we can re-write the equation with $y$ instead of $2^p$.

$m=\frac{1}{2} y(y-1)$

$2m=y^2-y$

$y^2-y-2m=0$

Use the quadratic formula to get:

$y = \frac{1+\sqrt{1+8m} }{2}$

or

$y = \frac{1-\sqrt{1+8m} }{2}$

Therefore, using natural log and log rules:

$2^p = \frac{1+\sqrt{1+8m} }{2}$, $ln(2^p)= ln(\frac{1+\sqrt{1+8m} }{2})$,$pln(2) = ln(\frac{1+\sqrt{1+8m} }{2})$, $p = \frac{ln(\frac{1+\sqrt{1+8m} }{2})}{ln(2)}$

or

$2^p = \frac{1-\sqrt{1+8m} }{2}$, $ln(2^p)= ln(\frac{1-\sqrt{1+8m} }{2})$, $pln(2) = ln(\frac{1-\sqrt{1+8m} }{2})$, $p = \frac{ln(\frac{1-\sqrt{1+8m} }{2})}{ln(2)}$

If I haven't made any mistakes this should be correct!