# log

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• Oct 18th 2008, 06:30 PM
franckherve1
log
1/log 25
25 what's its value?
• Oct 18th 2008, 06:31 PM
U-God
base e?
• Oct 18th 2008, 06:36 PM
franckherve1
it's like 25 at the base 1/log 25
• Oct 18th 2008, 06:41 PM
U-God
so like: $\displaystyle \frac{1}{Log_{25} 25}$ ?

Well what do you know about $\displaystyle Log_x x = y$
It's the same as: $\displaystyle x^y = x$ which implies x = 1.
• Oct 18th 2008, 06:42 PM
11rdc11
Quote:

Originally Posted by franckherve1
it's like 25 at the base 1/log 25

Does it look like this

$\displaystyle \frac{1}{\log{25}}$
• Oct 18th 2008, 06:46 PM
franckherve1
it's like 25 with exponent 1/log 25
• Oct 18th 2008, 06:53 PM
11rdc11
Quote:

Originally Posted by franckherve1
it's like 25 with exponent 1/log 25

do like U-god said then

so it simplfies to 1
• Oct 18th 2008, 07:03 PM
franckherve1
but the base is not 25 ...there's nothing so i guess the base should be 10
• Oct 18th 2008, 07:08 PM
U-God
In my country if there's no base, the base is taken to be e. But I'm aware that in other countries it's taken to be 10.

If it was 10 the answer would be $\displaystyle \frac{1}{1.398}$

However if it was e, the answer would be $\displaystyle \frac{1}{3.219}$

to 3 decimal places.
• Oct 18th 2008, 07:10 PM
11rdc11
Quote:

Originally Posted by franckherve1
but the base is not 25 ...there's nothing so i guess the base should be 10

so it is

$\displaystyle \frac{1}{\log{25^{25}}}$
• Oct 18th 2008, 07:13 PM
franckherve1
1
10
25
125
• Oct 18th 2008, 07:20 PM
franckherve1
1
10
25
125
• Oct 18th 2008, 07:28 PM
11rdc11
I think 125 but I'm still confused as to how the prob looks like
• Oct 18th 2008, 07:32 PM
franckherve1
it might be true but the question is how?
• Oct 18th 2008, 10:29 PM
CaptainBlack
Quote:

Originally Posted by franckherve1
1/log 25
25 what's its value?

Do you mean:

$\displaystyle 25^{1/\log(25)}$

If so, let the base be $\displaystyle b$ then:

$\displaystyle 25^{1/\log(25)}=(b^{\log(25)})^{1/\log(25)}=b$

CB
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