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Is it possible to solve logarithm equation when the base of the logarithm is unknown but the result is known. Here is an example:

$$ \log_{X} (\frac{223}{150}) = 20 $$

This basically means that if x multiplied 20 times we will get the fraction $\frac{223}{150}$. My guess is that this is the same as:

$$\sqrt[x]{\frac {223}{150}}$$

But it seems that it is incorrect. Appreciate any help, comment.

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  • $\begingroup$ It’s fine, though I’d prefer to write it $$\left(\frac{223}{150}\right)^{1/20}\;;$$ it turns out to be about $1.020024680802$. $\endgroup$ – Brian M. Scott Jun 27 '13 at 9:36
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    $\begingroup$ $\log_x a = \frac{\ln a}{\ln x}$ and now you have an equation you probably know how to solve for $x$. $\endgroup$ – Arthur Jun 27 '13 at 9:39
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This is basically the same as what Adam did, but the method is sometimes more straight-forward; so I will add it. There is a well-known identity:

$$\log_{x}{\frac{223}{150}}=\frac{\ln{\frac{223}{150}}}{\ln{x}}=20$$

Now we can separate the $x$, and solve the equation:

$$\Rightarrow \ln x = \frac{\ln{\frac{223}{150}}}{20}$$

$$\Rightarrow x=e^{\frac{\ln{\frac{223}{150}}}{20}}=\left( e^{\ln{\frac{223}{150}}} \right)^{\frac{1}{20}}=\sqrt[20]{\frac{223}{150}}$$

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$\log_X\frac{223}{150}=20$ means that $X^{20}=\frac{223}{150}$, so $X=\sqrt[20]\frac{223}{150}$

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Yes, $\log_x a$ denotes what you need to raise $x$ to the power of to get $a$. That is, if $\log_x a = b$, then $a = x^b$. In particular, $x = a^{1/b} = \sqrt[b]{a}$.

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To justify my comment, assume that changing base of logarithm is as simple as multiplying by some number, that is, that $$ \log_x a = y\ln a $$ holds for some $y$. Raising $x$ to the power of this equation, we get: $$ a = x^{y\ln a} = e^{\ln x \cdot y\ln a}= a^{y\ln x} $$ which means that $1 = y\ln x$, and we therefore have that $$ \log_x a = \frac{\ln a}{\ln x} $$ Putting this into your equation, we get $$ \frac{\ln \frac{223}{150}}{\ln x} = 20 \\\\ \ln x = 0.05\cdot \ln\frac{223}{150} \\\\ x = e^{\ln \frac{223}{150}/20}\\\\ x = \sqrt[20]{\frac{223}{150}} $$

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