If a raised to x= b raised to y = c raised to z and b^2 =ac, then show that y= 2×z/x+z
Answers
Step-by-step explanation:
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GIVEN THAT,
b {}^{2} = ac \: \: \: \: \: \: \: \: \: \: \: .....(1)b2=ac.....(1)
Also,
\begin{gathered}{a}^{x} = {b}^{y} = {c}^{z} \\ \\ let \: \\ {a}^{x} = {b}^{y} = {c}^{z} = p \\ \\ we \: have \\ {a}^{x} = p \\ \\ = > a = {p}^{ \frac{1}{x} } \: \\ \\ similarly \\ b = {p}^{ \frac{1}{y} } \: \: and \: \: c = {p}^{ \frac{1}{z} }\end{gathered}ax=by=czletax=by=cz=pwehaveax=p=>a=px1similarlyb=py1andc=pz1
Putting value of a, b and c in eq(1)
We get,
\begin{gathered}{( {p}^{ \frac{1}{y} }) }^{2} = {p}^{ \frac{1}{x} } \times {p}^{ \frac{1}{y} } \\ \\ {p}^{ \frac{2}{y} } = {p}^{ (\frac{1}{x} + \frac{1}{y} )}\end{gathered}(py1)2=px1×py1py2=p(x1+y1)
On equating the power of p
We get,
\begin{gathered}\frac{2}{y} = \frac{1}{x} + \frac{1}{z} \\ \\ \frac{2}{y} = \frac{z + x}{xz} \\ \\ \frac{1}{y} = \frac{x + z}{2xz} \\ \\ = > y = \frac{2xz}{x + z} \:\:\:\:\: [Proved] \:\end{gathered}y2=x1+z1y2=xzz+xy1=2xzx+z=>y=x+z2xz[Proved]