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|dw:1441998613082:dw|

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when we multiply same bases we should `add` exponents \[\huge\rm x^m \times x^n=x^{m+n}\] and when we divide same base , `subtract` their exponents \[\huge\rm \frac{ x^m }{ x^n }=x^{m-n}\]
so \[7^{-3} \times 7^{9}=?\]
|dw:1441998736665:dw|
7^-14
hmm how did you get *negative * 14
i just guess
don't guess it's easy when multiply same bases we should `add` their exponents so \[7^{-3} \times 7^{9}=\rm what ?\]
lol
-3+9=9-3
117649
well don't use calculator
here is an example \[\huge\rm 2^2 \times 2^3 = 2^{2+3} = 2^5\] base would stay the same \[\large\rm 7^{-3} \times 7^9=7^{-3+9}=7^?\]
2
mind telling how u got 2 ?
there 2 number 3 and 9
it so hard
i said just `add` exponents -3 and 9 are exponents 9-3= ??
9-3 = 6
yes right so numerator would be 7^6 \[\huge\rm \frac{ 7^6 }{ 7^{-8} }\] 2nd exponent rule when we divide same base we should `subtract` their exponents here is an example \[\huge\rm \frac{ 2^4 }{ 2^{-3} }=2^{4-(-3)}\]
2^4-(-3) you add
we would bec 2nd exponent is negative that would makes addition
4^2
what is 4^2 ?
|dw:1441999674369:dw|
here is an example \[\huge\rm \frac{ 2^4 }{ 2^{-3} }=2^{4-(-3)}=2^{4+3}=2^7\] basically we are subtracting 4 from -3 4-(-3) but bec 3 is negative that would makes positive 3
what about 4^2 ? \[\huge\rm \frac{ 7^6 }{ 7^{-8} } = ?\]
i think it A OR B
i have no idea what A or B ...
|dw:1441999981329:dw|
look at the example then solve this \[\huge\rm \frac{ 7^6 }{ 7^{-8} } = ?\]
ok
here is an example \[\huge\rm \frac{ 2^4 }{ 2^{-3} }=2^{4-(-3)}=2^{4+3}=2^7\] in this example base is 2
i said 2^7
you never said that anyways that's an example now solve ur question which is \[\huge\rm \frac{ 7^6 }{ 7^{-8} } = ?\]
move the exponent from denominator to top when you do that sign of the denominator exponent would change
ok ty
it was worng
which one ?
what did you get ?
i didn't give you the answer you didn't tell me which one you get
it was 7^14

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