how to do -log in head

[deleted357625]

Advertisement - Members don't see this ad

So i ran into a problem and I know how to do it, but I only had college algebra a year ago and forget.

-log(.04)

how do i do this without calculator I know log is base 10 or something, I'm just so use to using the calculator...

 

[UrshumMurshum]

So i ran into a problem and I know how to do it, but I only had college algebra a year ago and forget.

-log(.04)

how do i do this without calculator I know log is base 10 or something, I'm just so use to using the calculator...

First convert .04 to an actual number your head can use.
.04 = 4E-2

Absorb the minus sign
-log(4E-2) = log(1/(4E-2)) = log(1/4*E2) = log (1/4) + log(E2) = -log(4) + log(E2)

Which is about equal to -.5 + 2
which equals 3/2

Did you see how I approximated -log(4) to be 1/2. The square root of 10 is about 4 in my approximation and the square root gives the 1/2.

 

[kasho11]

I'm assuming this is mostly for pH calculations where you want to use this? If so just know that 10^-1 = pH 1 10^-2 = pH 2 10^-3 = pH 3 etc.

So if you have something like 4 x 10^-2, you know that it's between 10^-1 and 10^-2. It's a bit closer to 10^-2 so between 1 and 2, it will be closer to 2. 1.6 would be a very safe estimate.

 

[MedPR]

TBR teaches you to memorize that log3 is 0.47 and log2 is 0.3.

So -log(0.04) is -log(4*10^-2). Which is 2-log(4) which is 2-(log(2)+log(2)), which is 2-0.3-0.3 = 1.4.

 

[UrshumMurshum]

My way is better.

I provide no evidence for why.