Solving logarithmic equation $2\log(x) + 1 =\log(19x+2)$

Question

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

$$\log(x^2) + 1 = \log(19x +2)$$

$$\log(x^2) - \log(19x +2) = -1$$

$$\log\left(\frac {x^2} {19x +2}\right) = -1$$

Answer 1

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.

Answer 2

If you mean that $\log x=\ln x$, the solution is $$\log(\frac {x^2} {19x +2}) = -1$$ $$\frac {x^2} {19x +2}=e^{-1}$$ $$x^2-\frac{19}{e}x-\frac{2}{e}=0$$

Replace the $e$ with $10$ if you mean $\log x$ for base $10$. Then use the quadratic formula to solve it.

Answer 3

Remember how the logarithm is defined.

$$b^x=y \iff \log _b y = x$$

Therefore

$$\log \left ( \frac{x^2}{19x+2}\right)=-1$$

is equivalent to to the statement

$$\frac{x^2}{19x+2}=10^{-1}$$

which can be solved with the quadratic formula.