Solve logarithmic equation: $2\log_7 (x+2) - \log_7 (3x+10) = 0$

Question

Please, can someone check if this is the right answer $$x= -2 \pm \sqrt{3x + 10}$$

Thank you.

Answer 1

Just for checking answers, I highly recommend Wolfram Alpha: http://m.wolframalpha.com/input/?i=solve+2log_7+%28x+%2B+2%29+-+log_7+%283x+%2B+10%29+%3D+0&x=0&y=0

Answer 2

$$\begin{align} \log_7(x+2)^2-\log_7(3x+10) &= 0 \\ \\ \implies \log_7\frac{(x+2)^2}{(3x+10)} &= 0 \\ \\ \implies \frac{x^2+4x+4}{3x+10} &= 1 \\ \\ \implies x^2+4x+4 &= 3x+10 \\ \\ \implies x^2+x-6 &= 0 \end{align}$$

Solving for x will give you $x=-3$ and $x=2$. But only $2$ is acceptable as $\log_7(x+2)$ is not defined for $x=-3$.

Answer 3

Let's rewrite the equation. $$\begin{align} 2\log_7(x+2)-\log_7(3x+10) & = 0 \\ 2\log_7(x+2) & = \log_7(3x+10) \\ \log_7\left((x+2)^2\right) & = \log_7(3x+10) \\ \implies (x+2)^2 & = 3x+10 \\ \end{align}$$ Now all we have to do is solve for $x$ in that equation. $$\begin{align} (x+2)^2 & = 3x+10 \\ x^2+4x+4 & = 3x+10 \\ x^2+x-6 & = 0 \\ (x+3)(x-2) & = 0 \\ \end{align}$$ $$x=-3, \ 2$$ However only $x=2$ is a solution, because $2\log_7(x+2)$ is not defined for $x=-3$. Therefore the solution is: $$\color{green}{x=2}$$ Hope I helped