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I am new to logarithms and I am having trouble with this logarithm system. \begin{align*} \log_9(x) + \log_y(8) & = 2, \\ \log_x(9) + \log_8(y) & = 8/3. \end{align*} A step-by-step procedure would be highly appreciated.

Thanks in advance.

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    $\begingroup$ You haven't even had time to read the answer to your other logarithm question. Slow down, digest an answer before posing a new question. $\endgroup$ – Gerry Myerson Jul 13 '12 at 1:22
  • $\begingroup$ Please see here for how to typeset common math expressions with LaTeX. Here's a helpful trick: if you see a math expression on this site for which you want to know the LaTeX code, you can right click on it, go to "Show Math As", then choose "TeX Commands". $\endgroup$ – Zev Chonoles Jul 13 '12 at 1:23
  • $\begingroup$ Thanks, Zev. I didn't now about this. Very useful. $\endgroup$ – ncmathsadist Jul 13 '12 at 1:49
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Use the fact that $$\log_b(a) = \dfrac1{\log_a(b)}$$

Hence, if we denote $\log_9(x) = a$ and $\log_y(8) = b$, we get that \begin{align} a+b & = 2\\ \dfrac1a + \dfrac1b & = \dfrac83 \implies \dfrac{a+b}{ab} = \dfrac83 \implies ab = \dfrac34 \end{align} Now solve for $a$ and $b$ and hence $x$ and $y$.

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If $\log(\cdot)$ denotes logarithm in base $10,$ then $$\log_y(x) = \frac{\log(x)}{\log(y)}.$$ Apply through & simplify to see what system you will get. Substitute $X$ for $\log(x),$ and $Y$ for $\log(y).$ You should have 2 equations in 2 unknowns.

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