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If I have an equation, like this:

$k^{m} = m^{k}$,

I have been taught that I must apply natural logarithm, but others have taught me that I must apply the common logarithm.

Then, what should I use?

$m*ln(k)= k*ln(m)$, Natural logarithm (base $e$)?

$m*log(k)= k*log(m)$, Common logarithm (base $10$)?

I guess, the only difference is that:

Natural logarithm will leave the answer in terms of $e$ and common logarithm, will leave the answer in terms of $10$, but I'm not sure about this and I need a really good explanation, when it's convenient to use one and the other.

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    $\begingroup$ Always natural logarithm. Always. $\endgroup$ – Lord Shark the Unknown Jun 8 '18 at 15:20
  • $\begingroup$ $\ln$ is more popular than $\log$ $\endgroup$ – Sujit Bhattacharyya Jun 8 '18 at 15:23
  • $\begingroup$ Why ? @LordSharktheUnknown $\endgroup$ – Eduardo S. Jun 8 '18 at 15:28
  • $\begingroup$ Common and natural logs are proportional, so either will serve to answer the question. The natural logarithm is, well, natural (for mathematicians), It's the most useful in abstract mathematics. Base $10$ logarithms are a historical accident, coming from the fact that we have $10$ fingers, so base $10$ notation for integers. The common logarithm was invented to speed calculation. The only other logarithm you encounter these days is base $2$, useful in computer science. $\endgroup$ – Ethan Bolker Jun 8 '18 at 15:35
  • $\begingroup$ Use whatever logarithm you like. If you use base $k$ you get $m = k\log_k m$ and if you use base $m$ you get $k = m\log_m k$. But all $m \log_b k = k\log_b m$ will all be true statements, so whichever helps you solve will work. (Although frustratingly none really help much.) $\endgroup$ – fleablood Jun 8 '18 at 16:18
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$k^m = m^k \implies m\log_b k = k\log_a m$ will be true no matter what base $b$ you choose. And as $\log_b k$ an $\log_b m$ will always be in the same proportion ($\frac {\log_b k}{\log_b m} = \frac {\log_a k}{\log_a m}$ for all legitimate $a,b$) it doesn't matter which you pick.

Unless you are doing scientific notation where units and measurements are specifically designed to be represented in powers of $10$s there is nothing advantageous about $10$ over, say, $17$. And $k \log_{17} m = m\log_{17}k$ is a perfectly true and legitimate statement.

But if you are doing scientific notation where units are based on powers of $10$ then $\log_{10}$ has an obvious advantage.

If you are doing anything that might even remotely no matter how obliquely involve differentiation or integration (or even tangents or rates of change) you should use $\ln$ as it is ... natural. So that is why it is conventional to default to natural logs.

I'm surprised though that no-one has suggested logs based $k$ or $m$. That has the advantage of reducing an equation with two logarithms to one. $m = k\log_k m$ and $k = m\log_m k$ which could often help us. Although in this case it doesn't)

Use whatever base you like. Sometimes there will be practical advantages to use a specific base (Solve for $3^{27x} = y^{81} \cdot 3^{7}$ just screams for base $3$) but usually there won't be. The convention is math is base $e$. I imagine im must sciences is is also base $e$ but I imagine there are same instances where convention is $10$.

But it doesn't matter.

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$3^{27x} = y^{81} \cdot 3^{7}$

$\log_3 3^{27x} = \log_3(y^{81} \cdot 3^{7})$

$27x = 81\log_3 y + 7$

$x = 3\log_3 y + \frac 7{27}$.

But you could just as well (but not as easily solve it with natural logs.

$\ln 3^{27x} = \ln(y^{81} \cdot 3^{7})$

$27x \ln 3 = 81\ln y + 7\ln 3$.

$x = \frac {81\ln y + 7\ln 3}{27\ln 3}$

$= 3\frac {\ln y}{\ln 3} + \frac 7{27}$. The same answer.

And if we had use common $\log$ wed have gotten.

$x = 3\frac {\log y}{\log 3} + \frac 7{27}$.

All the same.

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  • $\begingroup$ thanks, good answer $\endgroup$ – Eduardo S. Jun 8 '18 at 22:42
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$$k^m=m^k\to m\log_a(k)=k\log_a(m)\to\frac1k\log_a(k)=\frac1m\log_a(m)\space\forall a\in \Bbb R^+$$

The most popular choice for $a$ is Euler's Constant $e$, for which we use the notation $\log_e(x)=\ln(x)$

Hence: $$\frac 1k\ln(k)=\frac 1m\ln (m)$$

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