pOH problem

[andyjl]

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i seem to have forgotten how to do this.

pOH = -log(3.2 x 10^-3) = 2.5

how do you get 2.5?

 

[poc91nc]

i seem to have forgotten how to do this.

pOH = -log(3.2 x 10^-3) = 2.5

how do you get 2.5?

pOH = 3 - log3.2 ~ 3 - log3 = 3 - 0.5 = 2.5

I posted this earlier... just remember these log values and you will be able to do most log calculations

log2 = 0.3
log3 = 0.5
log5 = 0.7
log7 = 0.8

 

[hoak]

-log (3.2 X 10^-3) = 3 - log3.2

since log3.2 ~ 0.5

=> 3 - 0.5 = 2.5

 

[andyjl]

pOH = 3 - log3.2 ~ 3 - log3 = 3 - 0.5 = 2.5

I posted this earlier... just remember these log values and you will be able to do most log calculations

log2 = 0.3
log3 = 0.5
log5 = 0.7
log7 = 0.8

what about the negative sign in front of the log.. don't the log rules say that you get rid of that by moving that up as an exponent to whatever you take the log to?

 

[Dotoday]

if you have a negative sign in front of the log, like -log2.5, then this is the same as log 1/(2.5). That is how you rewrite it. Then to get rid of log, you just put everythinh under the base of 10. So for example 3.5 = log1x, then 10^3.5 = 10^log1x, you get 10^3.5=1x. That is it, then you just solve for 10^3.5.

 

[levyusupov]

i seem to have forgotten how to do this.

pOH = -log(3.2 x 10^-3) = 2.5

how do you get 2.5?

the easiest way to do it is to look on the power the number is raised to. for example: the -log (1 X 10^-3)=3

the -log(3.2 X 10^-5)= app. 4.6

the -log (9 X 10^-7)= app. 6.1

if it simple such as -log1x10^-5, the answer will be thepower without the negative sign. but, if it is a number such as -log (6.7 X 10^-9), just know it will be LESS then 9 but bigger then 8, eyeball it. just go according to the power it is raised to, that is my advice.