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I have four constants $k_1$, $k_2$, $k_3$, $k_4$ and the following equation in an unknown $x$ (all are positive real):

$k_1k_2^x = k_3k_4^x$

How do I solve for x?

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Rearranging your equation yields $$ \frac {k_1}{k_3} = \left(\frac{k_4}{k_2}\right)^x, $$ and then taking logarithms of both sides gives $$ \log \frac {k_1}{k_3} = \log \left(\frac{k_4}{k_2}\right)^x = x \log \frac{k_4}{k_2} , $$ and so $$ x = \frac {\log \frac {k_1}{k_3}}{\log \frac {k_4}{k_2}} . $$

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Take the logarithm of both sides, to any base you like. We get $$\log k_1 +x\log k_2=\log k_3+x\log k_4.$$ Now we have a linear equation for $x$.

Remarks: $1.$ We have used two important facts about logarithms: (i) $\log(ab)=\log a+\log b$ and (ii) $\log(a^t)=t\log a$.

$2.$ Even though $k_1$, $k_2$, $k_3$ and $k_4$ are positive reals, this does not guarantee that $x$ will be positive.

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  • $\begingroup$ I've always found it unintuitive for some reason that $\log_c(a^b) = b\log_c(a)$. I find it hard to visualize a physical model that demonstrates this. $\endgroup$ – Andrew Tomazos Aug 25 '12 at 19:15
  • $\begingroup$ If $b=2$, $3$, $4$ then it just says $\log(a^2)=2\log a$, $\log(a63)=3\log a$, and so on, which can be thought of as coming from the $\log(xy)=\log x+\log y$. $\endgroup$ – André Nicolas Aug 25 '12 at 20:45
  • $\begingroup$ Wow, it seems obvious now - exponentiation becomes multiplication in log space - as multiplication becomes addition $\endgroup$ – Andrew Tomazos Aug 25 '12 at 21:09

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