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