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solve equations: $\log_x 10 +2\log_{10x} 10-3\log_{100x} 10=0$ so I tried to use $\log_a b=\frac{1}{\log_b a}$ but it didn't work for me.

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$$ \log_{10x} 10 = \frac{1}{\log_{10}(10x)} = \frac{1}{1+\log_{10} x} = \frac{1}{1+u} $$ and similarly $$ \log_{100x} 10 = \frac{1}{\log_{10}(100x)} = \frac{1}{2+\log_{10}x} = \frac{1}{2+u}. $$

So you have $$ \frac1u+ \frac{2}{1+u} - \frac{3}{2+u} = 0. $$ If you multiply both sides by $u(1+u)(2+u)$, you get $$ (1+u)(2+u) + 2u(2+u)-3u(1+u)=0. $$ Multiply that out, then collect like terms, then you have a quadratic equation.

( . . . and I just noticed that actually, it ends up being simpler than a quadratic equation because of a fortuitous cancelation.)

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Use $\log_a(b) = \frac{\log(b)}{\log(a)}$

Expand the equations put $\log(10x) = \log(10) + \log(x) = 1+\log(x)$ and $\log(100x) = \log(100) + \log(x) = 2+\log(x)$

You will get a quadratic in $\log(x)$ . Solve for that and find the value of $x$ .

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  • $\begingroup$ I changed $log$ to $\log$. This not only prevents italicization, but also provides proper spacing in expressions like $a\log b$. $\endgroup$ – Michael Hardy Dec 2 '13 at 17:26
  • $\begingroup$ Thanks for the info . I am new to using mathjax :) ! $\endgroup$ – abkds Dec 2 '13 at 17:28

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