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Why $\log_6 \sqrt{6} = 1/2$?
I know that $\sqrt{6} = 6^{1/2}$.

rubik
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user161304
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2 Answers2

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Since $\sqrt{6} = 6^{1/2}$ we have: $$\log_6 \sqrt{6} = \log_6 6^{1/2} = 1/2$$

You can think of $\log_6 n$ as "the exponent I need to apply to $6$ in order to obtain $n$". Well, since in this particular case $n=6^{1/2}$ it follows naturally that the exponent you need to apply is in fact $1/2$.

You can also evaluate the expression using these properties of logarithms: $$\begin{align*} \log_a b^c &= c \log_a b\\ \log_a a &= 1 \end{align*}$$

You can check that: $\log_6 6^{1/2} = \frac{1}{2} \log_6 6 = \frac{1}{2}$

rubik
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  • @user161304: Remember to accept the answer if it was helpful to you. Otherwise, don't hesitate to ask for clarifications. – rubik Jul 09 '14 at 16:35
  • So basically both of the 6's cancel out and leave you with 1/2 – user161304 Jul 09 '14 at 16:35
  • Well it's not very correct to say that they "cancel out", because actually they don't. You can think of $y = \log_6 6$ as $6^y = 6$. So what's the value of $y$, i.e. of the log? It can only be $1$. – rubik Jul 09 '14 at 16:37
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So what is a logarithm?

It is essentially the exponent value required, with a particular base, to give the value. In the equation below it is easy to see that x must 2. 6 is the base and x is the exponent.

$${6^x} = 36$$

The logarithm is the formal definition of this relationship.

$${Log_6(6^{1\over2})} = {x}$$

Let's rewrite that in the same style as the first:

$${6^x}={6^{1\over2}}$$

So obviously x must be $1\over2$. But if you notice the 6 is common to both sides and you really only care about the exponents, and that's where the log comes in.

Felix Castor
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