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?
 A: You will need the following identities: (For all $a,b,c>0$, $n$ real)
$$\begin{array}{l}(1) \hspace{10pt }\log_ab=x\Leftrightarrow a^x=b\\
(2) \hspace{10pt }\log_ab=\frac{\log_cb}{\log_ca}\\
(3) \hspace{10pt } n\log_ab=\log_a(b^n)\end{array}$$
Hence $\log_{\frac17}x^2=\frac{\log x^2}{\log \frac17}=\frac{\log x^2}{-\log 7}=-\log_7x^2=-2\log_7x$
and $\log_{49} x=\log_{7^2} x=\frac12\log_7x$
So $\log_7x+\log_{\frac17}x^2=-\log_7x$. Hence you have:
$$-\log_7x=\frac12\log_7x - 3 \hspace{5pt}\Rightarrow\hspace{5pt} \frac32\log_7x=3\hspace{5pt}\Rightarrow\hspace{5pt}\log_7x=2$$
A: Apply $\log_ab=\frac{\log b}{\log a}$ and $\log b^m=m\log b$
So,$\log_7x+\log_{\frac17}x^2=\log_{49}x-3$
becomes, $$\frac{\log x}{\log 7}-2\frac{\log x}{\log 7}=\frac{\log x}{2\log 7}-3$$
So, $\log_7x-2\log_7x=\frac12\log_7x-3,\implies \log_7x=2, x=7^2=49$
A: $$\log_{a}b=\dfrac{\log_c b}{ \log_c a}$$
$$log_7 x + log_{1/7} x^2 =log_49 x - 3 $$
So,
$$\log_{1/7}x^2=\dfrac{\log_7 x^2}{ \log_7 1/7}$$
$$\log_{1/7}x^2={-\log_7 x^2}$$
A: Even if you don't know properties of logs, you can procced as follows (this is the idea behind those proofs):
$$7^{\log_7 x + \log_{1/7} x^2} =7^{\log_{49} x - 3}$$
$$7^{\log_7 x} =x$$
$$7^{ \log_{1/7} x^2} = [\left( \frac{1}{7}\right)^{\log_{1/7} x^2}]^{-1}=x^{-2}$$
$$7^{\log_49 x - 3} = \sqrt{49^{\log_{49} x - 3}}=\sqrt{x-3}$$
A: $$\log_7 x + \log_{1/7} x^2 =\log_{7^2} (x) - 3 $$
$$\log_7 x + \frac{\log_7(1/7)}{\log_7 x^2} =\frac{\log_77^2}{\log_7 (x) }-3 $$
$$\log_7 x + \frac{-1}{\log_7 x^2} =\frac{2}{\log_7 (x) }-3 $$
$\log_7 x=t$ then we have
$$t-\frac{1}{t}=\frac{2}{t}-3$$
$$t^2+3t-3=0$$
$$\log_7x_1=\frac{-3+\sqrt{21}}{2},x_1=7^{\frac{-3+\sqrt{21}}{2}}$$
$$\log_7x_2=\frac{-3-\sqrt{21}}{2},x_2=7^{\frac{-3-\sqrt{21}}{2}}$$
