how do you solve
20*2^(13/s)

Question

how do you solve
20*2^(13/s)

anonymous · Answer

what? @Danielayolo

tkhunny · Answer

You don't.  It's not an equation.  Should there be "=" in there, somewhere?

anonymous · Answer

NORHING\

anonymous · Answer

20*2^(13/s)=y

anonymous · Answer

@tkhunny @Danielayolo @reemii

reemii · Answer

you want to find s as a function of y ?

reemii · Answer

do "as usual", at some point you will need to take the logarithm in base 2 on both sides.

tkhunny · Answer

$20\cdot 2^{13/s} = y$

Divide by 20

$2^{13/s} = \dfrac{y}{20}$

Now what?  I recommend a logarithm.

anonymous · Answer

@tkhunny like s=ln(8192)/ln(y)-ln(20)

tkhunny · Answer

No.  I have no idea how you managed that.  Please provide intermediate steps.

$\log\left(2^{13/s}\right) = \log(y/20)$

$\dfrac{13}{s}\log(2) = \log(y/20)$

Now what?

anonymous · Answer

@tkhunny remove the log?

anonymous · Answer

@reemii

reemii · Answer

remove $log_2(2)$? . it is equal to 1 so i think what you mean is correct.
then it's just a matter of isolating $s$ and obtaining an equation of the form $s=....$.
easy now.

anonymous · Answer

$$13/s=$$
idk how to do the other side :(

anonymous · Answer

@reemii

reemii · Answer

does not change.

anonymous · Answer

so its
$$13/s=\log(y/20)$$

reemii · Answer

yes, now you have $s=13/\log(y/20)$. you're done.

anonymous · Answer

@reemii how do i get an exact answer?

reemii · Answer

if yuo don't know the value of $y$, there's nothing else to do here.

tkhunny · Answer

I used Base 10 logs.  No removing ever should occur.