# Solve to get a value for $y.$

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.

• $y=\frac{log(c^2)}{log(b)}$ – poetasis Jun 23 at 12:22
• you have more equations than needed...$y=x.log(a)/log(b)=z.log(c)/log(b)$ – NoChance Jun 23 at 18:24

## 1 Answer

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 $$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$$

• I substituted obtain by get as I had to edit at least 6 characters. – JERRY_XLII Jun 23 at 16:02
• 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. – NoChance Jun 23 at 18:22
• The answer for y is $$y = \frac{2xz}{x+z}$$ – JERRY_XLII Jun 24 at 3:19