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I've been learning logarithm properties in a finite mathematics course and I am confused about a particular example in the text book. They are using the property $b^{log_b(x)}=x$, $x>0$, to solve the following problem:

$\frac{log_e(x)}{log_e(b)}$

  1. The first thing they do is replace x in the equation with $b^{log_b(x)}$ giving the result:

$\frac{log_e(b^{log_b(x)})}{log_e(b)}$

  1. Then use another property ($log_bM^p=p*log_bM$)to get:

$\frac{log_b(x)*log_e(b)}{log_e(b)}$

  1. The two "$log_e(b)$" cancel out and we are left with the answer

$log_b(x)$

My question is that before #3(or at any point really) I could replace the only "x" in the equation with blogb(x) again could I not? And I could continue replacing "x" with blogb(x) and get a massive equation. What I need to understand is why do they replace x with that logarithm and why can't I continue to replace it over and over(as pointless as that would be). I put in some numbers and constantly changing "x" to that log changes the result so it obviously can't be done continuously. Also in the property it states x>0. Does that mean that after using this property x now is greater than zero. Or can this property only be applied if x is greater than zero. In which case how can I know x=0 if no value has been assigned to x at this point?

I am not great at math and my teachers have told me this textbook has many typos and errors so I end up questioning everything. So any clarification on when I can and cannot use blogb(x)=x would be great thanks.

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4 Answers 4

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When $x$ and $b$ are positive, you can always say that $x = b^{\log_b(x)}$. (When $x$ is negative, this doesn't hold up since the logarithm of a negative number isn't well-defined in the reals).

Your book uses this because it is often useful to replace $x$ with $b^{\log_b(x)}$. One example of this property being useful is the proof you stated in your question. However, while you can always use this property (when $x$ and $b$ are both positive), it is not always useful.

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You can use any correct property whenever you want, as long as you apply it correctly. The real question is: when do you WANT to use it? The book replaces $x$ with $b^{\log_b(x)}$ only once because its sufficient for their purposes — after that, the expression simplifies to the desired result.

And I could continue replacing "$x$" with $b^{\log_b(x)}$ and get a massive equation.

Yes, you certainly could. But the end of this phrase answers your own question — you don't want to, because then, instead of getting something nice, you would get a "massive equation".

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You can always write $x=b^{\log_bx}$ when $x,b>0$. So that means that if you felt like it you could indeed write $$x=b^{\log_bx}=b^{\log_bb^{\log_bx}}=b^{\log_bb^{\log_bb^{\log_bx}}}=\ldots$$ However, such an expression seems like it probably wouldn't be very useful. On the other hand, writing $x=b^{\log_bx}$ is indeed useful, as evidenced by the very example you were working with. It actually becomes rather important in more advanced areas of mathematics - one example would be complex analysis, where $e^z$ and $\log z$ are defined for suitable complex numbers $z$, allowing us to define $$a^z=(e^{\log a})^z=e^{z\log a}.$$ Try coming up with another way to sensibly define, say, $\pi^{\sqrt{2}}$ to appreciate the power of this property.

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Your textbook is doing something weird, namely working from a specific application of the change of base formula to the "more frequently encountered expression" $\log_b x$. It is easy to derive the change of base formula to see what it is I am talking about.

Let $y=\log_b x$, whereby it must be true that $b^y=x$ by the definition of a logarithm. Now take the logarithm of base $a$ of both sides: $$ y=\log_b x\Leftrightarrow b^y=x\Leftrightarrow \log_a(b^y)=\log_a(x)\Leftrightarrow y\log_a(b)=\log_a(x). $$ From above, you can see that $y\log_a(b)=\log_a(x)$ or simply $y=\frac{\log_a(x)}{\log_a(b)}$. Remember that we defined $y=\log_b x$. That is, we have shown that $$ \log_b x=\frac{\log_a x}{\log_a b} $$ for any suitable base value $a$. Your specific problem is handled by letting $a=e$, where $e$ is Euler's constant.

Note that when $f(x)=a^x$ and $f^{-1}(x)=\log_a x$ we have $$ \log_a(a^x)=x,\qquad x\in\mathbb{R}\\[0.5em] a^{\log_a x}=x\qquad x>0. $$ Why? [Hint: This is because of the Inverse Function Property, and also consider the domain and range of $f$ as opposed to $f^{-1}\ldots$]

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