How do you solve this logarithmic equation?

Question

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.

Answer 1

I think you have got an error: $ x = log(y^x) + log(y) = log(y^x\cdot y) = log(y^{x+1})$

Answer 2

You've made an error.

$\log y^x +\log y$ is $\log y^{x+1}$, not $\log y^{2x}$

Answer 3

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}} $