Mathematics
OpenStudy (anonymous):

Solve. x=log[2](1/128)

OpenStudy (asnaseer):

first write 128 as a power of 2 - what do you get?

OpenStudy (anonymous):

163834

OpenStudy (anonymous):

i mean 16384

OpenStudy (asnaseer):

? - so you are saying that:$2^{16384}=128$

OpenStudy (asnaseer):

you need to find a value for n such that:$2^n=128$what is n?

OpenStudy (anonymous):

oh...

OpenStudy (anonymous):

7

OpenStudy (asnaseer):

correct, now can you use this result to express 1/128 as a power of 2? i.e. find n such that:$2^n=\frac{1}{128}$

OpenStudy (anonymous):

2^-1/7=1/128 ??

OpenStudy (asnaseer):

not quite, what you have found is:$2^{-1/7}=\frac{1}{\sqrt[7]{2}}$

OpenStudy (anonymous):

how??

OpenStudy (asnaseer):

$x^{\frac{1}{a}}=\sqrt[a]{x}$

OpenStudy (anonymous):

so the answer is 7square root of x?

OpenStudy (asnaseer):

therefore:$x^{-\frac{1}{a}}=\frac{1}{\sqrt[a]{x}}$

OpenStudy (asnaseer):

no, what I had asked you to do was to find n such that:$2^n=\frac{1}{128}$

OpenStudy (asnaseer):

you correctly found that:$2^7=128$

OpenStudy (anonymous):

ok...i think i get it now. Thanks for your help!!!

OpenStudy (asnaseer):

yw

OpenStudy (anonymous):

$\log_2\frac{ 1 }{ 128 }=x\\ (128)^{-1}=2^x\\ 2^{-7}=2^x\\\\ -7=x$