Is it possible to solve logarithm when base is unknown?

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.

• It’s fine, though I’d prefer to write it $$\left(\frac{223}{150}\right)^{1/20}\;;$$ it turns out to be about $1.020024680802$. – Brian M. Scott Jun 27 '13 at 9:36
• $\log_x a = \frac{\ln a}{\ln x}$ and now you have an equation you probably know how to solve for $x$. – Arthur Jun 27 '13 at 9:39

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:

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

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

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

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, $x = a^{1/b} = \sqrt[b]{a}$.

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 $$\frac{\ln \frac{223}{150}}{\ln x} = 20 \\\\ \ln x = 0.05\cdot \ln\frac{223}{150} \\\\ x = e^{\ln \frac{223}{150}/20}\\\\ x = \sqrt{\frac{223}{150}}$$