I'm stuck trying to solve
$$2\log(x) + 1 =\log(19x+2)$$
I know the solution has to be $x = 2$.
However I can't find the manual steps (Wolfram doesn't know the manual steps either).
This is all I got
$$\log(x^2) + 1 = \log(19x +2)$$
$$\log(x^2) – \log(19x +2) = -1$$
$$\log\left(\frac {x^2} {19x +2}\right) = -1$$
Best Answer
Hint: Write $-1$ as $\log\frac{1}{10}$. Then take antilogs both sides.
Edit: The solution: $$\frac{x^2}{19x+2}=\frac{1}{10}$$ Cross multiply: $$10x^2=19x+2$$ $$10x^2-19x-2=0$$ $$\left(x+\tfrac{1}{10}\right)(x-2)=0$$
There you have it! $x=2$. $x=-\frac{1}{10}$ gets rejected from your original question, but not from the one you solved. Think about it.