Help please! -log(2X10^(-4))

[lilietta2000]

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How do you calculate -log(2X10^(-4)) base 10 without using a calculator??

Thanks😕

 

[anonymuz]

this generally works for me... is approximate enough to get the right answer everytime for pH questions...

since its -log(2x10^-4)
i take 4 (from -4) then subtract .2 (from the 2) and so the final answer is
3.8 ish

so something like.... -log (7x10^-8) i would do
8 - .7 = 7.3ish.. its off by a little but the answers, at least when i was studying were very ranged so it was close enough.. im sure theres something better, but it worked for me

 

[Streetwolf]

How do you calculate -log(2X10^(-4)) base 10 without using a calculator??

Thanks😕

-log(2x10^-4) = -log(2) - log(10^-4)
= 4log(10) - log(2)
= 4 - log(2)

And log(2) is close to 0.3 so you get ~3.7 as an answer. You just have to know what log(2) is. You can guess well though.

If you don't know how I did all that stuff above, you should review your properties of logs.

 

[lilietta2000]

-log(2x10^-4) = -log(2) - log(10^-4)
= 4log(10) - log(2)
= 4 - log(2)

And log(2) is close to 0.3 so you get ~3.7 as an answer. You just have to know what log(2) is. You can guess well though.

If you don't know how I did all that stuff above, you should review your properties of logs.

How do you know log2 is about .3?...I guess I should memorize?

Thanks.

 

[rose786]

Here's a post I found to be very helpful:
http://forums.studentdoctor.net/showpost.php?p=2825247&postcount=7

 

[Streetwolf]

If you know log(2) = 0.3 then you know log(4) = 0.6 because log(4) = log(2^2) = 2*log(2).

You also know log(8) = 0.9 because log(8) = log(2^3) = 3*log(2).

You can get quite a few values with just one this way.

 

[lilietta2000]

Thanks everybody who helped🙂