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

Question

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!

---

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 1

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

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

Answer 2

There is nothing special about $2$ or $e$. For $x \gt 0$ and $a,b \gt 1$ you have $$\log_a(x)=\dfrac{\log_b(x)}{\log_b(a)} = \log_b(x) \cdot \log_a(b)$$

since $\displaystyle x= b^{\log_b(x)}$ and $\displaystyle \log_a(x) =\log_a \left(b^{\log_b(x)}\right)= \log_b(x) \cdot \log_a(b) $

Answer 3

$2^x=(e^{ln(2)})^x=e^{xln(2)}\rightarrow \ln(2^x) = x\ln2 \rightarrow x=\frac{\ln 2^x}{\ln2}$

Answer 4

In general, if you an equation of the form:$$a=b$$then whatever you apply to the left-hand-side of the equation, you also need to apply to the right-hand-side of the equation. e.g. you could multiply both sides by 2 to get:$$2a=2b$$ This also applies to functions, so you could apply the function f to both sides to get:$$f(a)=f(b)$$Now, going back to your original question where you had:$$2^x=129$$we can apply the log function to both sides to get:$$\log(2^x)=\log(129)$$there is no restriction on what the base of this has to be, as long as both sides use the same base. So, if you decided to take log base 2, then you would get:$$\log_2(2^x)=\log_2(129)$$$$\therefore x=\log_2(129)$$

Further clarification

Let $$b^x=y$$Now take logs of both sides (the base here is unimportant) to obtain: $$\log(b^x)=\log(y)$$ $$\therefore x\log(b)=\log(y)$$ $$\therefore x=\frac{\log(y)}{\log(b)}\tag{1}$$

Now look at the case when the logs that we took were with base $b$. This would result in:$$x=\frac{\log_b(y)}{\log_b(b)}=\frac{\log_b(y)}{1}=\log_b(y)$$

So taking the log to base $b$ is just a special case that makes the denominator of equation (1) equal to 1. Hope that makes sense.

Answer 5

You said you are comfortable with the following:

$2^{x}=129$

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

But where does that method come from?

$2^{x}=129$

$\implies \log_{2}2^x=\log_2 (129)$

$\implies x\log_{2}2=\log_2 (129)$

$\implies x=\log_2 (129)$

Answer 6

All logarithms are proportional to each other. Given $a,b>0$ we have $$ \log_a(y)=k\cdot\log_b(y) $$ for some constant factor $k>0$. This $k$ must simply be defined so that $a^k=b$ for then we see that $$ a^{k\cdot\log_b(y)}=b^{\log_b(y)}=y $$ as required. Now returning to your question, the factor $k$ cancels out whenever dividing logarithms: $$ \frac{\log_a(y)}{\log_a(z)}=\frac{\require{cancel}\cancel{k}\cdot\log_b(y)}{\cancel{k}\cdot\log_b(z)} $$ So in particular we get $$ \log_2(129)=\frac{\log_2(129)}{1}=\frac{\log_3(129)}{\log_3(2)}=\frac{\ln(129)}{\ln(2)}=\frac{\log_a(129)}{\log_a(2)} $$ where the second equality uses that $\log_2(2)=1$. In this way you should see that all the methods you have suggested should lead to the exact same answer! Also we see that any base $a$ would work the same.

Answer 7

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 $\log_4$ ? [...] How is it they all give the value of $x=3$ ?

But it does yield different values ! $$\boxed{\log_{\color{blue}a}(2^x)=\log_{\color{blue}a}(8){\color{red}\neq}\log_{\color{green}b}(2^x)=\log_{\color{green}b}(8).}$$ The value of the first pair is different than that of the second one: $$\boxed{x \log_{\color{blue}a}2=3\log_{\color{blue}a}2\ {\color{red}\neq}\ x \log_{\color{green}b}2=3\log_{\color{green}b}2}$$ It's just like asking, how come that $ax=ay$ yields the same result as $bx=by$ , namely that $x=y$ despite the fact that $a\neq b$, and that this holds true for all a, b $\neq0$.

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., $3$)?

If $A=B$ then $A+X=B+X$ , $A-X=B-X$ , $A\cdot X=B\cdot X$ , $A/X=B/X$ , $A^X=B^X$, $X/A=X/B$ , $X^A=X^B$ , $\sqrt[X]A=\sqrt[X]B$ , $\sqrt[A]X=\sqrt[B]X$ , $\log_XA=\log_XB$ , $\log_AX=\log_BX$, etc. What exactly is it that you find so puzzling or mysterious about it ?

Or perhaps you meant to ask why $\displaystyle{\frac{\log_{\color{red}x}a}{\log_{\color{red}x}b}=\log_ba,\quad\forall\ x\in(0,\infty)\setminus\{1\}}$ . Well, this is as obvious and uninteresting as asking, for instance, why $\displaystyle{\frac{x\cdot a}{x\cdot b}=\frac{a}{b},\quad\forall\ x\neq0}$ .

Answer 8

Consider that the logarithm is an operator which transform $a^x$ to $x \log(a)$. So, if you take the logarithm of both sides, the base does not matter and the result will always be the same. You could even use irrational base ($\pi$, $e$, ...).