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Negative exponent
- Thread starter usawhney
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usawhney said:(-8x^-4 4y^-4z)^-5 / (x^-4 y^7 z^-3) ^-1
Please help me i need it to know soon as possible i got exam tomorrow.
Thank you.
Negative exponents follow the same rules as the positive exponents.
You need to remember
\(\displaystyle a^{-m}\: = \: \frac{1}{a^m}\)
So you can convert the negative exponents to positive exponents and deal with those as usual.
Hello, usawhney!
I must assume that you know the rules for exponents.
\(\displaystyle \text{We have: }\;\frac{(\text{-}8)^{\text{-}5}\cdot(x^{\text{-}4})^{\text{-}5}\cdot(y^{\text{-}4})^{\text{-}5}\cdot(z)^{\text{-}5}} {(x^{\text{-}4})^{\text{-}1}\cdot(y^7)^{\text{-}1}\cdot(z^{\text{-}3})^{\text{-}1}}\)
. . . . . \(\displaystyle =\;\frac{(\text{-}8)^{\text{-}5}\cdot (x^{20})\cdot(y^{20})(z^{\text{-}5})} {(x^4)(y^{\text{-}7})(z^3)}\)
. . . . . \(\displaystyle =\;\frac{1}{(\text{-}8)^5} \cdot\frac{x^{20}}{x^4} \cdot\frac{y^{20}}{y^{\text{-}7}} \cdot\frac{z^{\text{-}5}}{z^3}\)
. . . . . \(\displaystyle =\;\frac{1}{\text{-}32,\!768}\cdot\frac{x^{16}}{1}\frac{y^{27}}{1}\cdot\frac{1}{z^8}\)
. . . . . \(\displaystyle =\;-\frac{x^{16}y^{27}}{32,\!768\, z^8}\)
I must assume that you know the rules for exponents.
\(\displaystyle \text{Simplify: }\;\frac{(\text{-}8\cdot x^{\text{-}4}\cdot y^{\text{-}4}\cdot z)^{\text{-}5}} {(x^{\text{-}4}\cdot y^7\cdot z^{\text{-}3}) ^{\text{-}1}}\)
\(\displaystyle \text{We have: }\;\frac{(\text{-}8)^{\text{-}5}\cdot(x^{\text{-}4})^{\text{-}5}\cdot(y^{\text{-}4})^{\text{-}5}\cdot(z)^{\text{-}5}} {(x^{\text{-}4})^{\text{-}1}\cdot(y^7)^{\text{-}1}\cdot(z^{\text{-}3})^{\text{-}1}}\)
. . . . . \(\displaystyle =\;\frac{(\text{-}8)^{\text{-}5}\cdot (x^{20})\cdot(y^{20})(z^{\text{-}5})} {(x^4)(y^{\text{-}7})(z^3)}\)
. . . . . \(\displaystyle =\;\frac{1}{(\text{-}8)^5} \cdot\frac{x^{20}}{x^4} \cdot\frac{y^{20}}{y^{\text{-}7}} \cdot\frac{z^{\text{-}5}}{z^3}\)
. . . . . \(\displaystyle =\;\frac{1}{\text{-}32,\!768}\cdot\frac{x^{16}}{1}\frac{y^{27}}{1}\cdot\frac{1}{z^8}\)
. . . . . \(\displaystyle =\;-\frac{x^{16}y^{27}}{32,\!768\, z^8}\)
Hi,
you kinda messed up in the equation..
(-8x[sup:3j5h1kcu]-4[/sup:3j5h1kcu]4y[sup:3j5h1kcu]-3[/sup:3j5h1kcu]z)[sup:3j5h1kcu]-5[/sup:3j5h1kcu]
(x[sup:3j5h1kcu]-4[/sup:3j5h1kcu]y[sup:3j5h1kcu]7[/sup:3j5h1kcu]z[sup:3j5h1kcu]-3[/sup:3j5h1kcu])[sup:3j5h1kcu]-1[/sup:3j5h1kcu]
The answer i got is :
x[sup:3j5h1kcu]16[/sup:3j5h1kcu]y[sup:3j5h1kcu]17[/sup:3j5h1kcu]
-33554432 z[sup:3j5h1kcu]8[/sup:3j5h1kcu]
It kind of looks wrong because of the long number.
you kinda messed up in the equation..
(-8x[sup:3j5h1kcu]-4[/sup:3j5h1kcu]4y[sup:3j5h1kcu]-3[/sup:3j5h1kcu]z)[sup:3j5h1kcu]-5[/sup:3j5h1kcu]
(x[sup:3j5h1kcu]-4[/sup:3j5h1kcu]y[sup:3j5h1kcu]7[/sup:3j5h1kcu]z[sup:3j5h1kcu]-3[/sup:3j5h1kcu])[sup:3j5h1kcu]-1[/sup:3j5h1kcu]
The answer i got is :
x[sup:3j5h1kcu]16[/sup:3j5h1kcu]y[sup:3j5h1kcu]17[/sup:3j5h1kcu]
-33554432 z[sup:3j5h1kcu]8[/sup:3j5h1kcu]
It kind of looks wrong because of the long number.
Hi usawhney,
there's a typo in there,
first can you check if you should have the "y term" to the power of -3 or -4,
then either way, you shouldn't end up with y[sup:sivlzgjn]17[/sup:sivlzgjn],
since (y[sup:sivlzgjn]-3[/sup:sivlzgjn])[sup:sivlzgjn]-5[/sup:sivlzgjn] is y[sup:sivlzgjn]15[/sup:sivlzgjn],
y[sup:sivlzgjn]15[/sup:sivlzgjn]/(y[sup:sivlzgjn]-7[/sup:sivlzgjn]) = y[sup:sivlzgjn]15+7[/sup:sivlzgjn]...
while (y[sup:sivlzgjn]-4[/sup:sivlzgjn])[sup:sivlzgjn]-5[/sup:sivlzgjn] divided by y[sup:sivlzgjn]-7[/sup:sivlzgjn] is y[sup:sivlzgjn]20[/sup:sivlzgjn]y[sup:sivlzgjn]7[/sup:sivlzgjn]=y[sup:sivlzgjn]27[/sup:sivlzgjn].
You'd get y[sup:sivlzgjn]17[/sup:sivlzgjn] if your numerator had a y[sup:sivlzgjn]-2[/sup:sivlzgjn] term.
there's a typo in there,
first can you check if you should have the "y term" to the power of -3 or -4,
then either way, you shouldn't end up with y[sup:sivlzgjn]17[/sup:sivlzgjn],
since (y[sup:sivlzgjn]-3[/sup:sivlzgjn])[sup:sivlzgjn]-5[/sup:sivlzgjn] is y[sup:sivlzgjn]15[/sup:sivlzgjn],
y[sup:sivlzgjn]15[/sup:sivlzgjn]/(y[sup:sivlzgjn]-7[/sup:sivlzgjn]) = y[sup:sivlzgjn]15+7[/sup:sivlzgjn]...
while (y[sup:sivlzgjn]-4[/sup:sivlzgjn])[sup:sivlzgjn]-5[/sup:sivlzgjn] divided by y[sup:sivlzgjn]-7[/sup:sivlzgjn] is y[sup:sivlzgjn]20[/sup:sivlzgjn]y[sup:sivlzgjn]7[/sup:sivlzgjn]=y[sup:sivlzgjn]27[/sup:sivlzgjn].
You'd get y[sup:sivlzgjn]17[/sup:sivlzgjn] if your numerator had a y[sup:sivlzgjn]-2[/sup:sivlzgjn] term.
They are always most awkward when you are getting used to them,
and when you do... the teacher throws in the kitchen sink as well.
Soroban gave the correct value for (-8)[sup:8ep8bgvb]-5[/sup:8ep8bgvb],
so check back on your work in calculating that.
Negative powers "represent" values in the denominator position,
indices are just fast to work with and convenient when you are used to them.
(-8)[sup:8ep8bgvb]-5[/sup:8ep8bgvb] is actually 1/[(-8)[sup:8ep8bgvb]5[/sup:8ep8bgvb]] which is -1/(8[sup:8ep8bgvb]5[/sup:8ep8bgvb])
since the exponent is odd.
and when you do... the teacher throws in the kitchen sink as well.
Soroban gave the correct value for (-8)[sup:8ep8bgvb]-5[/sup:8ep8bgvb],
so check back on your work in calculating that.
Negative powers "represent" values in the denominator position,
indices are just fast to work with and convenient when you are used to them.
(-8)[sup:8ep8bgvb]-5[/sup:8ep8bgvb] is actually 1/[(-8)[sup:8ep8bgvb]5[/sup:8ep8bgvb]] which is -1/(8[sup:8ep8bgvb]5[/sup:8ep8bgvb])
since the exponent is odd.