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

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I edited it to make it more readable. I don't think I missed anything, but if you think I did you can change it back. –  Amr Dec 9 '12 at 16:37
    
@Akos, is the base of log in the right hand side $49?$ –  lab bhattacharjee Dec 9 '12 at 16:45
    
The base is 49 on the right side. –  Ákos Kovács Dec 9 '12 at 16:47
    
I edited the question, now everything well formatted. –  Ákos Kovács Dec 9 '12 at 16:53

5 Answers 5

up vote 1 down vote accepted

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: $$-\log_7x=\frac12\log_7x - 3 \hspace{5pt}\Rightarrow\hspace{5pt} \frac32\log_7x=3\hspace{5pt}\Rightarrow\hspace{5pt}\log_7x=2$$

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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, $\log_7x-2\log_7x=\frac12\log_7x-3,\implies \log_7x=2, x=7^2=49$

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$$\log_{a}b=\dfrac{\log_c b}{ \log_c a}$$ $$log_7 x + log_{1/7} x^2 =log_49 x - 3 $$

So, $$\log_{1/7}x^2=\dfrac{\log_7 x^2}{ \log_7 1/7}$$ $$\log_{1/7}x^2={-\log_7 x^2}$$

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Even if you don't know properties of logs, you can procced as follows (this is the idea behind those proofs):

$$7^{\log_7 x + \log_{1/7} x^2} =7^{\log_{49} x - 3}$$

$$7^{\log_7 x} =x$$ $$7^{ \log_{1/7} x^2} = [\left( \frac{1}{7}\right)^{\log_{1/7} x^2}]^{-1}=x^{-2}$$ $$7^{\log_49 x - 3} = \sqrt{49^{\log_{49} x - 3}}=\sqrt{x-3}$$

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$$\log_7 x + \log_{1/7} x^2 =\log_{7^2} (x) - 3 $$

$$\log_7 x + \frac{\log_7(1/7)}{\log_7 x^2} =\frac{\log_77^2}{\log_7 (x) }-3 $$

$$\log_7 x + \frac{-1}{\log_7 x^2} =\frac{2}{\log_7 (x) }-3 $$

$\log_7 x=t$ then we have

$$t-\frac{1}{t}=\frac{2}{t}-3$$

$$t^2+3t-3=0$$ $$\log_7x_1=\frac{-3+\sqrt{21}}{2},x_1=7^{\frac{-3+\sqrt{21}}{2}}$$ $$\log_7x_2=\frac{-3-\sqrt{21}}{2},x_2=7^{\frac{-3-\sqrt{21}}{2}}$$

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