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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.

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    $\begingroup$ For example, because $2^x$ is defined as $\exp (x\ln 2)$. Also, $\log_2 y = \frac{\ln y}{\ln 2}$. $\endgroup$ – Daniel Fischer Nov 17 '13 at 22:19
  • $\begingroup$ They're taking log base 2 $\endgroup$ – Don Larynx Nov 17 '13 at 22:20
  • $\begingroup$ This is just another way to get $log_2(.)$. Suppose you have a calculator that does not have $log_2$ but has $\ln$ still you can calculate $log_2$. $\endgroup$ – hhsaffar Nov 17 '13 at 22:28
  • $\begingroup$ More over, if you have a calculator that does not have $\ln$ but has $log_2$ you can still compute $\ln$. Can you say how?(Well not exactly $\ln$ but pretty close if you know the value of $e$ to enough decimal points) $\endgroup$ – hhsaffar Nov 17 '13 at 22:30
  • $\begingroup$ Your edits have made your question make less sense than before. Any method whatsoever of solving $2^x = 8$ must give $x = 3$. This is because $2^3 = 8$ and $2^x \not= 8$ for any $x$ other than 3. $\endgroup$ – Rob Arthan Nov 19 '13 at 23:44
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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

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

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  • $\begingroup$ I cant thank you enough, you cleared up a lot of confusion, im sure some others have said what you in some form or another but this was the most easiest to understand. Thank you. $\endgroup$ – seeker Nov 20 '13 at 2:32
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    $\begingroup$ "Note that if you solve $2^y=e$ you get $y=ln(2)$." I don't think this is what you meant... $\endgroup$ – user64687 Nov 20 '13 at 10:12
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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) $

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  • $\begingroup$ Could you elaborate on this a little more, I feel you are saying something really pertinent to my question but I do not fully understand this, thank you. $\endgroup$ – seeker Nov 18 '13 at 19:20
  • $\begingroup$ Why did you substitute $$x = {b^{{{\log }_b}(x)}}$$ into the change of base formula? $\endgroup$ – seeker Nov 18 '13 at 19:37
  • $\begingroup$ @Assad - what I did was prove $\log_a(x) = \log_b(x) \cdot \log_a(b)$. Divide both sides by $\log_a(b)$ and you get $\dfrac{\log_a(x)}{\log_a(b)} = \log_b(x)$, which you call the "change of base formula". If you let $a=e$, $b=2$, and $x=129$ you get the result in your original question $\dfrac{\log_e(129)}{\log_e(2)} = \log_2(129)$, but this is more general than that $\endgroup$ – Henry Nov 18 '13 at 21:48
  • $\begingroup$ Okay I know this: $${\log _a}(x) = \frac{{{{\log }_b}(x)}}{{{{\log }_b}(a)}}$$ But what did you do or substitute to get: $$\eqalign{ & {\log _a}(x) = {\log _b}(x).{\log _a}(b) \cr & {\log _b}(x) = \frac{{{{\log }_a}(x)}}{{{{\log }_a}(b)}} \cr} $$ $\endgroup$ – seeker Nov 18 '13 at 22:54
  • $\begingroup$ Of your three expressions, the third is the same as the first, but with $a$ and $b$ swapped over, while the second is the same as the the third with both sides multiplied by $\log_a(b)$. The second also comes from taking base $a$ logarithms of $x=b^{\log_b(x)}$ $\endgroup$ – Henry Nov 19 '13 at 9:44
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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.

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  • $\begingroup$ A very clear answer. I see this, but my question is why is this so? A few of the other users from what I understand seem to be hinting at some kind proportionality, or scaling between logarithms or something, if you could elaborate on that a little, it would be appreciated. Thank you. $\endgroup$ – seeker Nov 18 '13 at 23:03
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$2^x=(e^{ln(2)})^x=e^{xln(2)}\rightarrow \ln(2^x) = x\ln2 \rightarrow x=\frac{\ln 2^x}{\ln2}$

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  • $\begingroup$ I am afraid I don't understand what you mean. $\endgroup$ – hhsaffar Nov 17 '13 at 22:48
  • $\begingroup$ Yeah I slaughtered the Latex $\endgroup$ – JimmyJackson Nov 17 '13 at 22:49
  • $\begingroup$ Your Implication is correct and I like the answer, but you left the reader to fill in one gap that I just wanted to point out $\endgroup$ – JimmyJackson Nov 17 '13 at 22:50
  • $\begingroup$ $2^{x}=(e^{ln(2)})^{x}=e^{xln(2)}=e^{ln(2^{x})}$ , the statement $e^{xln(2)}=e^{ln(2^{x})}$ implies $\ln(2^{x})=x\ln2$ (because we can take the natural log of both sides) which in turn implies $x=\frac{\ln2^{x}}{\ln2}$ $\endgroup$ – JimmyJackson Nov 17 '13 at 22:54
  • $\begingroup$ Actually, $x=\frac{\ln2^{x}}{\ln2}=\log_{2}(2^{x})$ this actually this is not too helpful to the question at hand. $\endgroup$ – JimmyJackson Nov 17 '13 at 23:14
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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.

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  • $\begingroup$ your answer confuses me a little, and im sure thats a short coming on my part. So let me get this right, you're basically saying a logarithm is scalable when you say this: $${\log _a}(y) = k.{\log _b}(y)$$ Could you explain this a little more? Why is this? $\endgroup$ – seeker Nov 18 '13 at 23:37
  • $\begingroup$ @Assad: I already gave in my answer an algebraic argument that with $a^k=b$ we get $a^{k\cdot\log_b(y)}=y$ showing that $k\cdot\log_b(y)$ works just like $\log_a(y)$ so they must be equal. $\endgroup$ – String Nov 19 '13 at 10:36
  • $\begingroup$ @Assad: But to give an example of how this is true, let us take $a=2$ and $b=32$. Then since $2^5=32$ we must have $k=5$. Then we should have $\log_2(1024)=5\cdot\log_{32}(1024)$ so here the example taken is $y=1024$. You may check that $\log_2(1024)=10$ and $\log_{32}(1024)=2$ so indeed $10=5\cdot 2$. Also we see that $2^{10}=2^{5\cdot 2}=32^2=1024$. $\endgroup$ – String Nov 19 '13 at 10:45
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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?

$2^{x}=129$

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

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

$\implies x=\log_2 (129)$

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  • $\begingroup$ That's an interesting way of looking at it, thanks! $\endgroup$ – seeker Nov 17 '13 at 23:12
  • $\begingroup$ But sure the first argument is how log is defined. From there you can prove stuff like $\log x^n=n\log x$. $\endgroup$ – JP McCarthy Nov 18 '13 at 0:16
  • $\begingroup$ What are you asking? $\endgroup$ – user66360 Nov 18 '13 at 2:38
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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 $\displaystyle{\frac{x\cdot a}{x\cdot b}=\frac{a}{b},\quad\forall\ x\neq0}$ .

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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, ...).

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