### 5x=40x^-1/2. Find x?

5x=40x^-1/2. Find x?

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timothy09
5 years ago
the real ad the corect aswer is 1/2. workigs: 5x=40x^-1/2. divide boths sides by 5 ad fid the reciprocal of the eqatio, sqare both sides, cross mltiply also to give.x^2¤8x=1. soo. mltiply both to give 8x^3=1. the. divide both by 8 ad cbe root both sides to give 1/3. soo. x=1/3.
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mayok2
5 years ago
x=4
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• Obipaul: Show Ur workings
5 years ago
lawal ibrahim
5 years ago
5x=40x*-1/2

multiply d powers by -2 so as 2 cancel out -1/2

5x*-2=40x*-1/2 x-2

5x*-2=40x

divide both sides by 5.we v

x*-2=8x.frm law of indices we v, 1/x*2=8x. cross multiply we v 1=8x*3.divide both side by 8 we v 1/8=x*3.findin d cube rut our ans=1/2
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ismac1726
5 years ago
X=4
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Kennedy01
5 years ago
5x=40x^(-1/2)

Divide both sides by(5)(x^-1/2)

5x/(5)(x^-1/2)=40x^-1/2/(5)(x^-1/2)

thus;

x/x^-1/2=8

from indices;

x^1-(-1/2)=8

=> x^1+1/2=8

x^3/2=8

multiply both exponents by 2/3;

x^(3/2*2/3)=8^(2/3)

Thus:

x=8^2/3

x=2^2

x=4
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timothy09
5 years ago
1/4
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berber
5 years ago
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HARDEYKUNLEY
5 years ago
4
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Samuel.nn
5 years ago
5x=40x^-1/2

divide through by 5

x=8x^-1/2

x=(8)*1/(x^1/2)

take the square of both sides of d equ.

x^2=64/x

cross multiply

x^3=64

take the cube root of both sides of d equation

x=4
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