algebra precalculus - Given $2^{x}=129$, why is it that I can use the natural logarithm to find $x$?

I've looked at an example in my textbook, it is:

$2^{x}=129$

$\ln \left( 2^{x}\right) =\ln \left( 129\right) $

$x\ln \left( 2\right) =\ln \left( 129\right) $

$ x=\dfrac {\ln \left( 129\right) }{\ln \left( 2\right) }$

My question is how is it that you can take logs to the base e and still obtain the right $x$? Shouldn't you have to take logs to base 2 to find the exponent $x$ that goes on 2 to give 129? Why is it that I can use the natural logarithm to find x in this instance? Also I've tried for other bases such as 3 as well and I get the right answer, why is this?

Furthermore when we usually solve an exponential function such as the above we would do:

$2^{x}=129$

$\log _{2}\left( 129\right) =x$

But how is it in this example they take logs of both sides?

I'm sorry if this doesn't make sense, I'm just very confused so take it easy on me, I'm studying at a pre-calculus mathematics level. Thank you for your help!

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EDIT: Okay I've opened a bounty on this question, partly because although i've received a lot of responses I still don't seem to understand this and hopefully someone else will come along with a fresh perspective or perhaps build on what others have wrote beforehand in a way that's conducive to my understanding. I hope this does not offend any of the people who have answered beforehand, its not your fault i cannot understand.

That said, what I would like to understand is the following:

(1) Why is it that if I have an equation $2^x=8$, that taking logs to any base b (where b>0) would always give me the same answer for x (i.e. 3)?:

$$\eqalign{ & {\log _2}({2^x}) = {\log _2}(8) \cr & {\log _3}({2^x}) = {\log _3}(8) \cr & {\log _4}({2^x}) = {\log _4}(8) \cr} $$

How is it they all give the value of $x=3$?

Shouldn't taking the equation $2^x=8$ to the base of say ${\log _2}$ give me a differing value to an equation taken to ${\log _4}$? So that leads me onto my next question:

(2) What property of logarithms does this? WHY is this so? I think from what I've gathered from here is it has to do with how one logarithm is proportional or scalable to another? But I don't understand how this applies, if someone could explain this I think i'd be able to understand.

I'd like any answers if possible to refer to the example I've already used, if possible. Thank you.

Answer

Note that if you solve $e^y=2$ you get $y=\ln(2)$.

This tells us that

$$2=e^{\ln(2)}$$

which you probably already know.

Now we can solve this equation two ways:

Method 1: as you solved it.

$$2^x=129 \Rightarrow x =\log_2(129) \,.$$

Method 2: Well, since $2=e^{\ln(2)}$ you can rewrite the equation as

$$(e^{\ln(2)})^x= 129 \,. $$

This is

$$e^{(x \ln (2))} =129 \,.$$

Since this is an exponential with base $e$, it suddenly makes sense to take the natural logarithm.

Then you get

$$x \ln(2) = \ln(129) \,.$$

Intuitively, this is the reason why you can take a different logarithm than the obvious one, in a hidden way you use that any number $a$ can be rewritten as $b^{\log_b(a)}$ and this way any exponential with basis $a$ becomes an exponential with basis $b$. This process is called " the change of base for logarithm formula", and it can be done much faster by the simple formula

$$\log_a(b)=\frac{\log_c(b)}{\log_c(a)} \,.$$