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

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lilietta2000

Dr.lili
10+ Year Member
15+ Year Member
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How do you calculate -log(2X10^(-4)) base 10 without using a calculator??

Thanks😕
 
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 :laugh:
 
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.
 
-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.
 
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