Problem Calc. Logs Plz Help!

[Saternoc]

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Hello how would I solve this specific problem by hand?

-log [square root of (1x10^-5)]

I know the answer is 2.5 but what is the easiest way to solve for this?

Also for future references how would I calculate something like this by hand:

square of (2 X10^-5) ?

 

[UCB05]

Hello how would I solve this specific problem by hand?

-log [square root of (1x10^-5)]

I know the answer is 2.5 but what is the easiest way to solve for this?

Also for future references how would I calculate something like this by hand:

square of (2 X10^-5) ?

With odd powers, I rewrite it to the next closest even power:
1x10^-5 = 10x10^-6
Which makes taking the sqrt of the exponential easier, and you can then break that up into factors.
sqrt(10x10^-6) = sqrt(10)xsqrt(10^-6) = ~3.2x10^-3

The next part takes some logic.
log (1x10^-3) = -3. Now, as the number increases, it gets closer to 10x10^-3, or 1x10^-2. The log of that is -2. This means that log (3.2x10^-3) is somewhere between -3 and -2. I would estimate about -2.5. -log(..etc..) = 2.5

squaring scientific notation is very simple. Remember the basic (associative?) law of multiplication.
(a x b)(a x b) = (a x a x b x b)
(2 x 10^-5)^2 = (2 x 2 x 10^-5 x 10^-5) = 4 x 10^-10

 

[hausee]

With odd powers, I rewrite it to the next closest even power:
1x10^-5 = 10x10^-6
Which makes taking the sqrt of the exponential easier, and you can then break that up into factors.
sqrt(10x10^-6) = sqrt(10)xsqrt(10^-6) = ~3.2x10^-3
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-log(3.2x10^-3)=-(log3.2+log10^-3)=-log3.2-log10^-3=-0.5-(-3)=2.5

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log(axb)=loga+logb
log(a/b)=loga-logb