Solution for $\log_7x+\log_{\frac17}x^2=\log_{49}x-3$ What is the right solution for $\log_7x+\log_{\frac17}x^2=\log_{49}x-3$. What logarithm identities used?

$$\log_7 x + \log_{1/7} x^2 =\log_{7^2} (x) - 3 $$
$$\log_7 x + \frac{\log_7(1/7)}{\log_7 x^2} =\frac{\log_77^2}{\log_7 (x) }-3 $$
$$\log_7 x + \frac{-1}{\log_7 x^2} =\frac{2}{\log_7 (x) }-3 $$
$\log_7 x=t$ then we have
$$t-\frac{1}{t}=\frac{2}{t}-3$$
$$t^2+3t-3=0$$
$$\log_7x_1=\frac{-3+\sqrt{21}}{2},x_1=7^{\frac{-3+\sqrt{21}}{2}}$$
$$\log_7x_2=\frac{-3-\sqrt{21}}{2},x_2=7^{\frac{-3-\sqrt{21}}{2}}$$
