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Please help me solve this question. Solve for y.

$1.\quad a^x=b^y=c^z$

$2.\quad b^2=ac$

I figured out that $b^{2+y}=a^{x+1}c^{z+1}$, but I am not able to go further, please help me.

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  • $\begingroup$ $y=\frac{log(c^2)}{log(b)}$ $\endgroup$ – poetasis Jun 23 at 12:22
  • $\begingroup$ you have more equations than needed...$y=x.log(a)/log(b)=z.log(c)/log(b)$ $\endgroup$ – NoChance Jun 23 at 18:24
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Substituting $$a^x=b^y=c^z=t$$ so we get (using $b^2=ac$) $t^{2/y}=t^{1/x+1/z}$ and we get $$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$$

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  • $\begingroup$ I substituted obtain by get as I had to edit at least 6 characters. $\endgroup$ – JERRY_XLII Jun 23 at 16:02
  • $\begingroup$ What if $x=2, b=1, y=1, c=1, z=1$.... The LHS in the last equation is $2$ but the RHS is $1.5$! Please explain. $\endgroup$ – NoChance Jun 23 at 18:22
  • $\begingroup$ The answer for y is $$y = \frac{2xz}{x+z}$$ $\endgroup$ – JERRY_XLII Jun 24 at 3:19

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