Is anyone else having problems solving logs and square roots without a caluculat

[JMC117]

calculator? I know how to do the general chemistry problem but i always get it wrong when it comes to estimating this numbers. does anyone have any tips?

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[Streetwolf]

Square roots: Figure out what 2 numbers the answer is between. Do this by locating the closest perfect square above and below the # you are at.

Ex: sqrt(116) is between 10 (sqrt 100) and 11 (sqrt 121).

Next, see how close you are to these numbers. In general if you are in the middle area between the two perfect squares I would make a guess around halfway between the two numbers. If it's getting close to either end, I would guess around those numbers.

Here since we have 116 which is much closer to 121 than to 100, I would make a guess of about 10.8. The answer is about 10.77 so I was pretty damn close.

If I was asking for the sqrt(110) I would guess somewhere near 10.5 because that's right up the middle.

Logs: First bring everything down to log(any # between 1 and 10 not counting those), example log(2.4), log(9.99), log(0.02), log(5).

How can you do this? Use properties of logs.

You'll probably have a problem like this:

-log(4.6 * 10^-3)

Okay so first property: log(A * B) = log(A) + log(B)
Second property: log(A/B) = log(A) - log(B)
Third property: log(A^n) = n*log(A)

You'll probably only see the first one. You can use the third one too but you don't have to.

Now obviously you want to use it and get -log(4.6) + -log(10^-3).

The second part of that is +3 (you should know this).

So you have 3 - log(4.6). That's what you want: a log with a # between 1 and 10.

Now remember that log(1) = 0 and that log(10) = 1. This means we'll be subtracting a number between 0 and 1, so our answer is between 2 and 3. This will probably narrow you down to at most 2 answers 95% of the time.

Can we do better? Yeah! Why? Because we're always between 1 and 10 so there's always a solid rule you can apply. Remember the number 3.2. This is the square root of 10 (approx). So log(3.2) = 1/2. If you are left with any log between 1 and 3.2, you will be adding/subtracting a number LESS THAN 1/2. If you have the log of any number between 3.2 and 10, you will be adding/subtracting a number GREATER THAN 1/2. So like the above example with square roots, you can have 3 cases:

log of a number closer to 1
log of a number closer to 3.2
log of a number closer to 10

So here's 3 examples:

4 - log(7.6) will be closer to 3. I'd guess about 3.2. Answer is 3.12. Okay not spot on but that will give you your answer 99.99% of the time.

3 - log(3.6) will be close to 2.5, slightly LESS. I'd guess 2.45. The answer is 2.44.

6 + log(1.9) will be closer to 6 and ABOVE 6 (remember we're adding here). I'd guess 6.3. The answer is 6.28.

==

Just remember that since you don't have a calculator, they don't expect you to get it directly on the nose. That means they won't have answers like 6.4, 6.45, 6.48, 6.51. They'll have something like 4.56, 5.12, 5.67, 6.12. For either of these two methods you will be able to narrow it down to within 1 integer so at worst you'd have to choose between 5.12 and 5.67.

You want to always FIRST find the integer the answer is within. Then see if there's only 1 answer choice for that. If there are multiple answers within that integer range, you need to do the other stuff - guessing which integer it is closer to.

 

[grldent]

Square roots: Figure out what 2 numbers the answer is between. Do this by locating the closest perfect square above and below the # you are at.

Ex: sqrt(116) is between 10 (sqrt 100) and 11 (sqrt 121).

Next, see how close you are to these numbers. In general if you are in the middle area between the two perfect squares I would make a guess around halfway between the two numbers. If it's getting close to either end, I would guess around those numbers.

Here since we have 116 which is much closer to 121 than to 100, I would make a guess of about 10.8. The answer is about 10.77 so I was pretty damn close.

If I was asking for the sqrt(110) I would guess somewhere near 10.5 because that's right up the middle.

Logs: First bring everything down to log(any # between 1 and 10 not counting those), example log(2.4), log(9.99), log(0.02), log(5).

How can you do this? Use properties of logs.

You'll probably have a problem like this:

-log(4.6 * 10^-3)

Okay so first property: log(A * B) = log(A) + log(B)
Second property: log(A/B) = log(A) - log(B)
Third property: log(A^n) = n*log(A)

You'll probably only see the first one. You can use the third one too but you don't have to.

Now obviously you want to use it and get -log(4.6) + -log(10^-3).

The second part of that is +3 (you should know this).

So you have 3 - log(4.6). That's what you want: a log with a # between 1 and 10.

Now remember that log(1) = 0 and that log(10) = 1. This means we'll be subtracting a number between 0 and 1, so our answer is between 2 and 3. This will probably narrow you down to at most 2 answers 95% of the time.

Can we do better? Yeah! Why? Because we're always between 1 and 10 so there's always a solid rule you can apply. Remember the number 3.2. This is the square root of 10 (approx). So log(3.2) = 1/2. If you are left with any log between 1 and 3.2, you will be adding/subtracting a number LESS THAN 1/2. If you have the log of any number between 3.2 and 10, you will be adding/subtracting a number GREATER THAN 1/2. So like the above example with square roots, you can have 3 cases:

log of a number closer to 1
log of a number closer to 3.2
log of a number closer to 10

So here's 3 examples:

4 - log(7.6) will be closer to 3. I'd guess about 3.2. Answer is 3.12. Okay not spot on but that will give you your answer 99.99% of the time.

3 - log(3.6) will be close to 2.5, slightly LESS. I'd guess 2.45. The answer is 2.44.

6 + log(1.9) will be closer to 6 and ABOVE 6 (remember we're adding here). I'd guess 6.3. The answer is 6.28.

==

Just remember that since you don't have a calculator, they don't expect you to get it directly on the nose. That means they won't have answers like 6.4, 6.45, 6.48, 6.51. They'll have something like 4.56, 5.12, 5.67, 6.12. For either of these two methods you will be able to narrow it down to within 1 integer so at worst you'd have to choose between 5.12 and 5.67.

You want to always FIRST find the integer the answer is within. Then see if there's only 1 answer choice for that. If there are multiple answers within that integer range, you need to do the other stuff - guessing which integer it is closer to.

wow this is the best way someone has every explained me. Thanks.

 

[Streetwolf]

Oh one more thing I forgot to mention on logs...

If you had log(132.8 * 10^-9), you don't want to make this log(132.8) + log(10^-9) because then you'd be stuck with log(132.8).

You want to do log(1.328 * 10^-7) which then becomes log(1.328) + log(10^-7) and so on.

That way you have your log(# between 1 and 10).

 

[JMC117]

Square roots: Figure out what 2 numbers the answer is between. Do this by locating the closest perfect square above and below the # you are at.

Ex: sqrt(116) is between 10 (sqrt 100) and 11 (sqrt 121).

Next, see how close you are to these numbers. In general if you are in the middle area between the two perfect squares I would make a guess around halfway between the two numbers. If it's getting close to either end, I would guess around those numbers.

Here since we have 116 which is much closer to 121 than to 100, I would make a guess of about 10.8. The answer is about 10.77 so I was pretty damn close.

If I was asking for the sqrt(110) I would guess somewhere near 10.5 because that's right up the middle.

Logs: First bring everything down to log(any # between 1 and 10 not counting those), example log(2.4), log(9.99), log(0.02), log(5).

How can you do this? Use properties of logs.

You'll probably have a problem like this:

-log(4.6 * 10^-3)

Okay so first property: log(A * B) = log(A) + log(B)
Second property: log(A/B) = log(A) - log(B)
Third property: log(A^n) = n*log(A)

You'll probably only see the first one. You can use the third one too but you don't have to.

Now obviously you want to use it and get -log(4.6) + -log(10^-3).

The second part of that is +3 (you should know this).

So you have 3 - log(4.6). That's what you want: a log with a # between 1 and 10.

Now remember that log(1) = 0 and that log(10) = 1. This means we'll be subtracting a number between 0 and 1, so our answer is between 2 and 3. This will probably narrow you down to at most 2 answers 95% of the time.

Can we do better? Yeah! Why? Because we're always between 1 and 10 so there's always a solid rule you can apply. Remember the number 3.2. This is the square root of 10 (approx). So log(3.2) = 1/2. If you are left with any log between 1 and 3.2, you will be adding/subtracting a number LESS THAN 1/2. If you have the log of any number between 3.2 and 10, you will be adding/subtracting a number GREATER THAN 1/2. So like the above example with square roots, you can have 3 cases:

log of a number closer to 1
log of a number closer to 3.2
log of a number closer to 10

So here's 3 examples:

4 - log(7.6) will be closer to 3. I'd guess about 3.2. Answer is 3.12. Okay not spot on but that will give you your answer 99.99% of the time.

3 - log(3.6) will be close to 2.5, slightly LESS. I'd guess 2.45. The answer is 2.44.

6 + log(1.9) will be closer to 6 and ABOVE 6 (remember we're adding here). I'd guess 6.3. The answer is 6.28.

==

Just remember that since you don't have a calculator, they don't expect you to get it directly on the nose. That means they won't have answers like 6.4, 6.45, 6.48, 6.51. They'll have something like 4.56, 5.12, 5.67, 6.12. For either of these two methods you will be able to narrow it down to within 1 integer so at worst you'd have to choose between 5.12 and 5.67.

You want to always FIRST find the integer the answer is within. Then see if there's only 1 answer choice for that. If there are multiple answers within that integer range, you need to do the other stuff - guessing which integer it is closer to.

thanks you so much!! this is amazing