Prove that :
xa(b−c)xb(a−c)÷(xbxa)c\dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^cxb(a−c)xa(b−c)÷(xaxb)c = 1
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Solving L.H.S. of the given equation :
⇒xa(b−c)xb(a−c)÷(xbxa)c=xab−acxab−bc÷xbcxac=xab−ac−(ab−bc)÷xbc−ac=xab−ab−ac+bc÷xbc−ac=xbc−ac÷xbc−ac=xbc−acxbc−ac=1.\Rightarrow \dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c = \dfrac{x^{ab - ac}}{x^{ab - bc}} ÷ \dfrac{x^{bc}}{x^{ac}} \\[1em] = x^{ab - ac - (ab - bc)} ÷ x^{bc - ac} \\[1em] = x^{ab - ab - ac + bc} ÷ x^{bc - ac} \\[1em] = x^{bc - ac} ÷ x^{bc - ac} \\[1em] = \dfrac{x^{bc - ac}}{x^{bc - ac}} \\[1em] = 1.⇒xb(a−c)xa(b−c)÷(xaxb)c=xab−bcxab−ac÷xacxbc=xab−ac−(ab−bc)÷xbc−ac=xab−ab−ac+bc÷xbc−ac=xbc−ac÷xbc−ac=xbc−acxbc−ac=1.
Since, L.H.S. = R.H.S. = 1.
Hence, proved that xa(b−c)xb(a−c)÷(xbxa)c=1\dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c = 1xb(a−c)xa(b−c)÷(xaxb)c=1.
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Solve for x and y, if :
(32)x÷2y+1=1 and 8y−164−x2=0(\sqrt{32})^x ÷ 2^{y + 1} = 1 \text{ and } 8^y - 16^{4 - \dfrac{x}{2}} = 0(32)x÷2y+1=1 and 8y−164−2x=0
(xaxb)a+b−c(xbxc)b+c−a(xcxa)c+a−b\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}(xbxa)a+b−c(xcxb)b+c−a(xaxc)c+a−b = 1
If ax = b, by = c and cz = a, prove that :
xyz = 1.
If ax = by = cz and b2 = ac, prove that :
y = 2xzx+z\dfrac{2xz}{x + z}x+z2xz.
