Answer:
Step-by-step explanation:
Assuming you're solving for p:
[tex]m=2^{-1} *2^p(2^p-1)[/tex]
Let [tex]y=2^p[/tex]
Now we can re-write the equation with [tex]y[/tex] instead of [tex]2^p[/tex].
[tex]m=\frac{1}{2} y(y-1)[/tex]
[tex]2m=y^2-y[/tex]
[tex]y^2-y-2m=0[/tex]
Use the quadratic formula to get:
[tex]y = \frac{1+\sqrt{1+8m} }{2}[/tex]
or
[tex]y = \frac{1-\sqrt{1+8m} }{2}[/tex]
Therefore, using natural log and log rules:
[tex]2^p = \frac{1+\sqrt{1+8m} }{2}[/tex], [tex]ln(2^p)= ln(\frac{1+\sqrt{1+8m} }{2})[/tex],[tex]pln(2) = ln(\frac{1+\sqrt{1+8m} }{2})[/tex], [tex]p = \frac{ln(\frac{1+\sqrt{1+8m} }{2})}{ln(2)}[/tex]
or
[tex]2^p = \frac{1-\sqrt{1+8m} }{2}[/tex], [tex]ln(2^p)= ln(\frac{1-\sqrt{1+8m} }{2})[/tex], [tex]pln(2) = ln(\frac{1-\sqrt{1+8m} }{2})[/tex], [tex]p = \frac{ln(\frac{1-\sqrt{1+8m} }{2})}{ln(2)}[/tex]
If I haven't made any mistakes this should be correct!
