How to extract the base of a logarithm as unknown? ( Logarithm equation)

Question

In order to extract the base of a logarithm as unknown, I am given the following formula

If $\log_x({n})$=m then $x=$ $b^\frac{\log_b{n}}{m}$.

How to explain this formula? Does it actually work with any base $b$. What are the conditions imposed on $b$?

Would it also work in case the number $n$ were expressed in terms of $x$, I mean, in case the unknown $x$ were also present in the argument of the log function ( on the LHS)?

For example, how could I solve , with this formula, an equation such as :

$\log_x{2x}=100$?

Thanks for your help.

Answer 1

$log_x(n)=m \Rightarrow x^m=n$ $\Rightarrow (x^m)^\frac{1}{m}=n^\frac{1}{m}$ $\Rightarrow x=n^\frac{1}{m}$

$x=b^{log_b(n)^\frac{1}{m}}$ $\Rightarrow x=b^{\frac{1}{m}log_b(n)}$

And b and x must follow conditions required for base of logarithms. a>0 and $a\ne1$

Answer 2

As is well-known,

$$\log_x(n)=\frac{\log(n)}{\log(x)}$$ where $\log$ denotes the natural logarithm or any other.

Then

$$\log(x)=\frac{\log(n)}m$$ and

$$x=e^{\log(n)/m}=b^{\log_b(n)/m}.$$

Answer 3

$$\log_x n=m \implies \frac{\ln n}{\ln x}=m \implies \ln x=\frac{\ln n}{m} \implies x=e^{\frac{\ln n}{m}}$$ Yes it works for all positice values of $x$ and $m$.

Answer 4

As for the question on how the formula works:
From the given equation, we have $$x^m=n$$ $$\Rightarrow x=n^{\frac{1}{m}}$$ Since we know any number $k$ can be written as $$b^{\log_b ({k})}$$ (With the condition imposed on $b$ being the same as the ones imposed on the base of the logarithm such as the number can neither be negative nor $0$ or $1$) So from the above value of x we have : $$x=b^{\log_b (n^{\frac{1}{m}})}$$ $$\Rightarrow x= b^{\frac{\log_b (n)}{m}}$$ As for the second question:
We can write : $$x=3^{\frac{\log_3(2x)}{100}}$$ $$\Rightarrow x=(2x)^{\frac{1}{100}}$$ $$\Rightarrow x^{100}= 2x$$ $$\Rightarrow x=2^{\frac{1}{99}}$$ (As $x$ cannot be $0$ because it is the input and the base of the logarithm in the question)