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What is the value of the following expression?

$$\log_b \left( m^{\log_b n} \right)$$

As far as I know it should be:

$$\log_bn\;\times\;\log_bm$$

Can it be simplified further? If so how?

Also, I read somewhere that it is equal to:

$$\log_b m$$

If that is correct, how is it possible?

Also, if it matters, the place where I read about the second answer, used base $10$.

Here's the image of the book where I found it:

Image

Farhan Anam
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  • Is it (log (m ^ log n)) or (log m) ^ (log n) ? – babbupandey Feb 12 '16 at 19:31
  • It's the the former - log (m ^ log n) – Farhan Anam Feb 12 '16 at 19:33
  • Yeah, I can't figure out what's going on in the second rectangle either... but I do get that $\log u=0$ from the first rectangle. If you do as well, then don't worry about it. – Bobson Dugnutt Feb 12 '16 at 20:25
  • @Lovsovs How do you get $log;u = 0$ from the first rectangle? I'm not that experienced with logarithms.. Please help me and thanks for your time. – Farhan Anam Feb 12 '16 at 20:34
  • I've added the explaination now. You need to know the log-rule $\log \left( \frac{u}{v} \right)=\log u - \log v$. I hope that helps. I'm just going to shamelessly put this out there, but if you'd like, you can "accept" my answer (the tick under the vote up/down-arrows) and make me very happy! :D – Bobson Dugnutt Feb 12 '16 at 20:57

2 Answers2

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I'm assuming you mean $\log_b\left( m^{\log_bn} \right).$

Then $\log_bn\;\times\;\log_bm$ is correct, and that cannot be simplified any further. Perhaps you are thinking about $\log_b(n \times m)=\log_bn+\log_bm$?

And it cannot equal $\log_bm$ - where would the $n$ have gone?

EDIT to explain recently introduced picture in question:

$\log u=\log\frac{z}{y} \log x+\log\frac{x}{z} \log y+\log\frac{y}{x} \log z=(\log z-\log y)\log x+(\log x-\log z)\log y+(\log y-\log x)\log z=\log z \log x-\log y\log x+\log x\log y-\log z\log y+\log y\log z-\log x\log z=0$

Mithlesh Upadhyay
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Bobson Dugnutt
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Wow, I don't like that book at all. It is true that $$ \log u = \log x^{\log (z/y)} + \log y^{\log(x/z)} + \log z^{\log(y/x)} $$ implies $$ \log u = \log \frac zy + \log \frac xz + \log \frac yx, $$ because the right-hand formulas in both equations are equal to zero, but since what we need to do is to to show that the right-hand side of the first equation is zero, we can't legitimately use that fact to argue this implication.

It certainly is not true that in general $\log x^a + \log y^b + \log z^b = a + b + c,$ so I really can't see how that step of the solution was justified.

I think a better approach is never to write the expression $x^{\log (z/y)} y^{\log(x/z)} z^{\log(y/x)}$ at all. Instead, go back one step in the chain of equations $(1)$ in the book's "solution", where we see the equal expression $x^{\log z - \log y} y^{\log x - \log z} z^{\log y - \log x}$. That's an easier expression to work with. Set \begin{align} u &= x^{\log z - \log y} y^{\log x - \log z} z^{\log y - \log x}, \\ \log u &= \log \left(x^{\log z - \log y} y^{\log x - \log z} z^{\log y - \log x}\right) \\ &= \log \left(x^{\log z - \log y}\right) + \log\left(y^{\log x - \log z}\right) + \log\left(z^{\log y - \log x}\right) \\ &= (\log z - \log y)\log x + (\log x - \log z)\log y + (\log y - \log x)\log z \\ &= \log z \log x - \log y \log x + \log x \log y - \log z \log y + \log y \log z - \log x \log z \\ &= 0. \end{align}

David K
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  • The book is good otherwise but this one came as a surprise. – Farhan Anam Feb 12 '16 at 21:12
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    OK, even in a good book sometimes something bad sneaks past the author and the editors. If it only happens once or twice, that's what errata and corrected editions are for. – David K Feb 12 '16 at 21:15