Thread: Log question. (without a calc evaluate)

1. 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

2. 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:

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

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

$y = \log_5(125)$

This means that $5^y = 125$

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

So now you have:

$\log_2(\log_3(3))$

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

$x = \log_3(3)$

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

This gives:

$\log_2(1)$

Set it equal to a random variable:

$z = \log_2(1)$

$2^z = 1$

Clearly $z = 0$

3. Originally Posted by Mush
Err doesn't really make sense.

Do you mean:

$\log_2(\log_3(\log_5(125)))$?
Indeed I do, though there is no front bracket '(' on there 125 for my question.
it says: Without a calculator, evaluate the following:
$\log_2(\log_3(\log_5125)))$

4. 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.

5. Thanks a bunch Mush, very quick response.

so its just 0(1(3)))

6. Originally Posted by captain_jonno
Thanks a bunch Mush, very quick response.

so its just 0(1(3)))
Nah it's more like:

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

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

$= \log_2(1)$

$= 0$

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