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Mathematics 16 Online
OpenStudy (anonymous):

\(y = a^{-x} \) \(y = (\frac{1}{a} )^x\) both above graphs are same, plz help understand the exponent property.,.

terenzreignz (terenzreignz):

You know that \[\huge n^{-p}=\frac{1}{n^{p}}\] ?

OpenStudy (anonymous):

yes i knw, but im confused about why that works

terenzreignz (terenzreignz):

you want me to prove it?

OpenStudy (anonymous):

looking at graph of \(a^{-x}\) and \(a^x\) i see that both are reflections of each other in in y-axis

OpenStudy (anonymous):

proof or some reasoning that i can understand is ok...

terenzreignz (terenzreignz):

Well, first you have to know that \[\huge m^{a}m^{b}=m^{a+b}\]

terenzreignz (terenzreignz):

You have that, right?

OpenStudy (anonymous):

ya that one im ok wid it

terenzreignz (terenzreignz):

Ok, so... \[\huge n^{0}=1\] right?

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

well 0 = p-p, so... \[\huge n^{p-p}=1\]

terenzreignz (terenzreignz):

right?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

how to get to ratio

terenzreignz (terenzreignz):

Then, from our property, \[\huge n^{p}n^{-p}=1\] right?

OpenStudy (anonymous):

wow ! so \(n^{-p}\) is multiplicative inverse of \(n^p\) which equals to \(\frac{1}{n^p}\)

terenzreignz (terenzreignz):

Yes, that's right :)

terenzreignz (terenzreignz):

You get it now?

OpenStudy (anonymous):

i get it thank you so soooo much xD

terenzreignz (terenzreignz):

No problem :)

OpenStudy (anonymous):

turning to this site when i get stuck is much productive than wasting time googling and watching youtube videos hoping my question wil be answered there. appreciate your time very much... . :you guys rock ! xD , ty

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