# How do you solve this logarithmic equation?

While reading through my textbook, I came across this particular equation: $$x = x\log (y) + \log (y)$$ But they solve it by doing this: $$x = x\log (y) + \log (y)$$ $$x = (x + 1)\log(y)$$ $$\frac {x}{x+1} = \log(y)$$ $$y = 10^{\frac{x}{x+1}}$$ Which is fine, but I don't understand why they didn't do it like this: $$x = x\log (y) + \log (y)$$ $$x=\log(y^x)+\log(y)$$ $$x=\log(y^{2x})$$ $$x=2x\log(y)$$ $$\frac{x}{2x}=\log(y)$$ $$y=10^{\frac{x}{2x}}$$ I'm confused over which one is correct.

• Yep, I feel like an idiot! Thanks, shost71 and MPW! – dr_nefario Mar 16 '15 at 2:34

I think you have got an error: $x = log(y^x) + log(y) = log(y^x\cdot y) = log(y^{x+1})$
$\log y^x +\log y$ is $\log y^{x+1}$, not $\log y^{2x}$
When you do $$log(y^x) + log(y)$$ this equals: $$log(y^x \cdot y) = log(y^{x+1})$$
which gives $y = 10^{\frac{x}{x+1}}$