show that x^a(b-c)/x^b(a-c)÷(x^b/x^a)^c=1
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{ x^[a(b - c)] / x^[b(a - c)] } / (x^b / x^a)^c = 1
If so, then the first thing that I'll do is simplify the exponents. Distributive properties in the first two and the exponent of an exponent is the product of the exponents in the third:
[ x^(ab - ac) / x^(ab - bc) ] / [x^(bc) / x^(ac)] = 1
Division of fractions is the same as the multiplication of the reciprocal:
[ x^(ab - ac) / x^(ab - bc) ] * [x^(ac) / x^(bc)] = 1
In each set of braces, the quotient of two numbers of the same base is the same as the differences in the exponents:
x^[(ab - ac) - (ab - bc)] * [x^(ac - bc)] = 1
and simplify:
x^(ab - ac - ab + bc) * x^(ac - bc) = 1
x^(-ac + bc) * x^(ac - bc) = 1
And finally, the product of two numbers of the same base is the same as the sum of the exponents:
x^[(-ac + bc) + (ac - bc)] = 1
and simplify:
x^(-ac + bc + ac - bc) = 1
x^(0) = 1
Anything to the 0 power is 1, so this is TRUE.
Hope This Helps :)
If so, then the first thing that I'll do is simplify the exponents. Distributive properties in the first two and the exponent of an exponent is the product of the exponents in the third:
[ x^(ab - ac) / x^(ab - bc) ] / [x^(bc) / x^(ac)] = 1
Division of fractions is the same as the multiplication of the reciprocal:
[ x^(ab - ac) / x^(ab - bc) ] * [x^(ac) / x^(bc)] = 1
In each set of braces, the quotient of two numbers of the same base is the same as the differences in the exponents:
x^[(ab - ac) - (ab - bc)] * [x^(ac - bc)] = 1
and simplify:
x^(ab - ac - ab + bc) * x^(ac - bc) = 1
x^(-ac + bc) * x^(ac - bc) = 1
And finally, the product of two numbers of the same base is the same as the sum of the exponents:
x^[(-ac + bc) + (ac - bc)] = 1
and simplify:
x^(-ac + bc + ac - bc) = 1
x^(0) = 1
Anything to the 0 power is 1, so this is TRUE.
Hope This Helps :)
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273
here is the solution....
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