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Question Number 72496 by Shamim last updated on 29/Oct/19
if a^x =b, b^y =c, c^z =a then prove that,  xyz=0.
$$\mathrm{if}\:\mathrm{a}^{\mathrm{x}} =\mathrm{b},\:\mathrm{b}^{\mathrm{y}} =\mathrm{c},\:\mathrm{c}^{\mathrm{z}} =\mathrm{a}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}, \\ $$$$\mathrm{xyz}=\mathrm{0}. \\ $$
Answered by som(math1967) last updated on 29/Oct/19
a^x =b ⇒a^(xy) =b^y   a^(xy) =c⇒a^(xyz) =c^z   ∴a^(xyz) =a ∴xyz=1   so I think it should be xyz=1
$${a}^{{x}} ={b}\:\Rightarrow{a}^{{xy}} ={b}^{{y}} \\ $$$${a}^{{xy}} ={c}\Rightarrow{a}^{{xyz}} ={c}^{{z}} \\ $$$$\therefore{a}^{{xyz}} ={a}\:\therefore{xyz}=\mathrm{1}\: \\ $$$${so}\:{I}\:{think}\:{it}\:{should}\:{be}\:{xyz}=\mathrm{1} \\ $$
Answered by Tanmay chaudhury last updated on 29/Oct/19
a=c^z   a=(b^y )^z =b^(yz)   a=(a^x )^(yz) =a^(xyz) →so xyz=1
$${a}={c}^{{z}} \\ $$$${a}=\left({b}^{{y}} \right)^{{z}} ={b}^{{yz}} \\ $$$${a}=\left({a}^{{x}} \right)^{{yz}} ={a}^{{xyz}} \rightarrow{so}\:{xyz}=\mathrm{1} \\ $$

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