Thread: Log question. (without a calc evaluate)

Hi there,

I have a question, how do you evaluate the following without a calculator?

log2(log3(log5^125)))

Thanks

ps. log2, log3 and log5 are the bases

Originally Posted by captain_jonno

Hi there,

I have a question, how do you evaluate the following without a calculator?

log2(log3(log5^125)))

Thanks

ps. log2, log3 and log5 are the bases

Err doesn't really make sense.

Do you mean:

$\displaystyle \log_2(\log_3(\log_5(125))) $?

If so, take the very first term, and set it equal to y:

$\displaystyle y = \log_5(125) $

This means that $\displaystyle 5^y = 125 $

You should be able to tell that $\displaystyle 125 = 5^3 $, and hence $\displaystyle y = 3 $.

So now you have:

$\displaystyle \log_2(\log_3(3)) $

Now take that term again, and set it equal to another variable, x.

$\displaystyle x = \log_3(3) $

$\displaystyle 3^x = 3 $. Clearly $\displaystyle x = 1$

This gives:

$\displaystyle \log_2(1) $

Set it equal to a random variable:

$\displaystyle z = \log_2(1) $

$\displaystyle 2^z = 1 $

Clearly $\displaystyle z = 0 $

Originally Posted by Mush

Err doesn't really make sense.

Do you mean:

$\displaystyle \log_2(\log_3(\log_5(125))) $?

Indeed I do, though there is no front bracket '(' on there 125 for my question.
it says: Without a calculator, evaluate the following:
$\displaystyle \log_2(\log_3(\log_5125))) $

Originally Posted by captain_jonno

Indeed I do, though there is no front bracket '(' on there 125 for my question.

See my edited post above.

Thanks a bunch Mush, very quick response.

so its just 0(1(3)))

Originally Posted by captain_jonno

Thanks a bunch Mush, very quick response.

so its just 0(1(3)))

Nah it's more like:

$\displaystyle \log_2(\log_3(\log_5125))) $

$\displaystyle = \log_2(\log_3(3)) $

$\displaystyle = \log_2(1) $

$\displaystyle = 0 $

Thanks, thats actually how i wrote it down in the end . good stuff