Bet u can’t solve this

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!

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