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Given that $(ab)^2=(bc)^4=(ca)^x=abc$ Then what is the value of $x$?

$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$

Then I am lost, any other easier way to solve?

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$a,b,c$ are positive reals and $abc\ne 1$ –  miosaki Nov 28 '13 at 13:27

2 Answers 2

up vote 2 down vote accepted

Taking logarithm gives: $$2(\log a+\log b)=4(\log b+\log c)=x(\log c+\log a)=\log a+\log b+\log c$$

then taking $2(\log a+\log b)=\log a+\log b+\log c$

and $4(\log b+\log c)=\log a+\log b+\log c$

we would get $\log a+\log b-\log c=0$ & $3\log b+3\log c-\log a=0$ and by solving these two equations we get $\log b=-\log c$

similarly $\log a=-\log b$ then the solution becomes obvious......as $x=\frac12$

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Hint: Split these two equalities into

$$(ab)^2 = (ca)^x$$ and $$(bc)^4 = (ca)^x$$

Then use $\log$ on both equations and see what happens ;-)

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