need help with square root problems

[eric beans]

Asked to put these in order a =√5 ; b = 2.5 ; c = log10 400 from smallest to largest.

How would you tackle this without a calculator?

My thinking was I don't know what the decimal number for √5 is so I thought well it's between √4 and √9 so I thought it's probably around 2.1 or so (pretty small compared to 2.5. But when I got to log10 400, I was not certain where to go. I know it's between log10 100 and log10 1000 but not sure if it's bigger or smaller than 2.5 since it's pretty close to the middle. Any thoughts?

 

[Farzin]

It is easy to see that [MATH]b^2=6.25[/MATH] and [MATH]a^2=5[/MATH] therefore [MATH]b>a[/MATH]But for the [MATH]c[/MATH], there is a trick that can help you to find the approximate logarithm of any number with up to one decimal digit.
Here is how we do it:
[MATH]A)[/MATH] If the first digit of our number is [MATH]1[/MATH], we count the number of digits and subtract one unit, then we put a decimal point and write [MATH]2[/MATH]Example:
[MATH]log_{10}16730248[/MATH], there are [MATH]8[/MATH] digits in this number therefore we subtract one unit and write [MATH]7.[/MATH] and because the first digit is [MATH]1[/MATH] then we put [MATH]2[/MATH] after the decimal point, now the answer is [MATH]log_{10}16730248≈7.2[/MATH]
[MATH]B)[/MATH] If the first digit of our number is [MATH]2[/MATH] to [MATH]7[/MATH], we do the same as above but this time we add [MATH]2 units[/MATH] to the first digit and write it after the decimal point.
Example:
[MATH]log_{10}65347[/MATH], there are [MATH]5[/MATH] digits in this number therefore we write [MATH]4.[/MATH] and add [MATH]2 units[/MATH] to it's first digit which is [MATH]6[/MATH] and the answer is [MATH]log_{10}65347≈4.8[/MATH]
[MATH]C)[/MATH] If the first digit of our number is [MATH]8[/MATH] or [MATH]9[/MATH], we do the same as above but this time we put [MATH]9[/MATH] after the decimal point.
Example:
[MATH]log_{10}8546732[/MATH], there are [MATH]7[/MATH] digits in this number therefore we write [MATH]6.[/MATH] and put [MATH]9[/MATH] after the decimal point and the answer is [MATH]log_{10}8546732≈6.9[/MATH]
Now for [MATH]log_{10}400[/MATH] there are 3 digits in [MATH]400[/MATH] therefore we subtract one unit and because the first digit is from [MATH]2[/MATH] to [MATH]7[/MATH] we have to add [MATH]2units[/MATH] to it's first digit and the answer is [MATH]log_{10}400≈2.6[/MATH]thus the answer to your problem is [MATH]c>b>a[/MATH]

 

[Deleted member 4993]

Asked to put these in order a =√5 ; b = 2.5 ; c = log10 400 from smallest to largest.

How would you tackle this without a calculator?

My thinking was I don't know what the decimal number for √5 is so I thought well it's between √4 and √9 so I thought it's probably around 2.1 or so (pretty small compared to 2.5. But when I got to log10 400, I was not certain where to go. I know it's between log10 100 and log10 1000 but not sure if it's bigger or smaller than 2.5 since it's pretty close to the middle. Any thoughts?

Another way to resolve the size factor between b and c would be to look at

10^2.5 = 10² * 10^0.5 < 10² * 16^0.5

 

[eric beans]

Another way to resolve the size factor between b and c would be to look at

10^2.5 = 10² * 10^0.5 < 10² * 16^0.5

I don't get it. where did the 16^0.5 come from? What happened to the log10 400?

 

[eric beans]

It is easy to see that [MATH]b^2=6.25[/MATH] and [MATH]a^2=5[/MATH] therefore [MATH]b>a[/MATH]But for the [MATH]c[/MATH], there is a trick that can help you to find the approximate logarithm of any number with up to one decimal digit.
Here is how we do it:
[MATH]A)[/MATH] If the first digit of our number is [MATH]1[/MATH], we count the number of digits and subtract one unit, then we put a decimal point and write [MATH]2[/MATH]Example:
[MATH]log_{10}16730248[/MATH], there are [MATH]8[/MATH] digits in this number therefore we subtract one unit and write [MATH]7.[/MATH] and because the first digit is [MATH]1[/MATH] then we put [MATH]2[/MATH] after the decimal point, now the answer is [MATH]log_{10}16730248≈7.2[/MATH]
[MATH]B)[/MATH] If the first digit of our number is [MATH]2[/MATH] to [MATH]7[/MATH], we do the same as above but this time we add [MATH]2 units[/MATH] to the first digit and write it after the decimal point.
Example:
[MATH]log_{10}65347[/MATH], there are [MATH]5[/MATH] digits in this number therefore we write [MATH]4.[/MATH] and add [MATH]2 units[/MATH] to it's first digit which is [MATH]6[/MATH] and the answer is [MATH]log_{10}65347≈4.8[/MATH]
[MATH]C)[/MATH] If the first digit of our number is [MATH]8[/MATH] or [MATH]9[/MATH], we do the same as above but this time we put [MATH]9[/MATH] after the decimal point.
Example:
[MATH]log_{10}8546732[/MATH], there are [MATH]7[/MATH] digits in this number therefore we write [MATH]6.[/MATH] and put [MATH]9[/MATH] after the decimal point and the answer is [MATH]log_{10}8546732≈6.9[/MATH]
Now for [MATH]log_{10}400[/MATH] there are 3 digits in [MATH]400[/MATH] therefore we subtract one unit and because the first digit is from [MATH]2[/MATH] to [MATH]7[/MATH] we have to add [MATH]2units[/MATH] to it's first digit and the answer is [MATH]log_{10}400≈2.6[/MATH]thus the answer to your problem is [MATH]c>b>a[/MATH]

Does this only work for log10?

 

[Farzin]

Yes it only works for Logarithm with base 10

 

[Deleted member 4993]

I don't get it. where did the 16^0.5 come from? What happened to the log10 400?

Let

log₁₀400 = x

Then:

400 = 10^x ............................................ (1)

We have shown:

10^2.5 = 10² * 10^0.5 < 10² * 16^0.5

10^2.5 < 10² * 16^0.5

10^2.5 < 400 .................. then using (1)

10^2.5 < 10^x

2.5 < x

get it now?

Work with pencil and paper instead of staring at the screen......

 

[Steven G]

Another way to resolve the size factor between b and c would be to look at

10^2.5 = 10² * 10^0.5 < 10² * 16^0.5

Sweet. I guess I taught you well!

 

[eric beans]

Let

log₁₀400 = x

Then:

400 = 10^x ............................................ (1)

We have shown:

10^2.5 = 10² * 10^0.5 < 10² * 16^0.5

10^2.5 < 10² * 16^0.5

10^2.5 < 400 .................. then using (1)

10^2.5 < 10^x

2.5 < x

get it now?

Work with pencil and paper instead of staring at the screen......

you misunderstand. that's not what the original post was asking. at the time i didn't know about the log rules for converting to decimal numbers. THAT was the problem. obviously after knowing the conversion rules, it's plainly obvious even without using that extra series of superfluous time consuming steps.

 

[Steven G]

but how/where did you get "10^2 * 16^0.5" without calculator? where did 16^0.5 come from? If i knew 10^2 * 16^0.5 was equal to 400, I wouldn't have needed to ask the question in the first place. but i didn't.

10^2 = 100. It just happens to turn out that 100*4= 400 and 4 is sqrt(16). Also, since 10<16 we have 10^.5 < 16^.5

10^2.5 = 10²*10^.5 <10²*16^.5 = 100*4=400.

Then it follows that 2.5 < log₁₀400. This is because if we say that log₁₀400 = 2.5 then it follows that 10^2.5 =400. This is not true. That power of 10 must be larger than 2.5!

 

[eric beans]

10^2 = 100. It just happens to turn out that 100*4= 400 and 4 is sqrt(16). Also, since 10<16 we have 10^.5 < 16^.5

10^2.5 = 10²*10^.5 <10²*16^.5 = 100*4=400.

Then it follows that 2.5 < log₁₀400. This is because if we say that log₁₀400 = 2.5 then it follows that 10^2.5 =400. This is not true. That power of 10 must be larger than 2.5!

Ah! That is far clearer. I know understand. I guess I was thrown by the ^0.5 which is actually another way of writing squareroot.

 

[eric beans]

It is easy to see that [MATH]b^2=6.25[/MATH] and [MATH]a^2=5[/MATH] therefore [MATH]b>a[/MATH]But for the [MATH]c[/MATH], there is a trick that can help you to find the approximate logarithm of any number with up to one decimal digit.
Here is how we do it:
[MATH]A)[/MATH] If the first digit of our number is [MATH]1[/MATH], we count the number of digits and subtract one unit, then we put a decimal point and write [MATH]2[/MATH]Example:
[MATH]log_{10}16730248[/MATH], there are [MATH]8[/MATH] digits in this number therefore we subtract one unit and write [MATH]7.[/MATH] and because the first digit is [MATH]1[/MATH] then we put [MATH]2[/MATH] after the decimal point, now the answer is [MATH]log_{10}16730248≈7.2[/MATH]
[MATH]B)[/MATH] If the first digit of our number is [MATH]2[/MATH] to [MATH]7[/MATH], we do the same as above but this time we add [MATH]2 units[/MATH] to the first digit and write it after the decimal point.
Example:
[MATH]log_{10}65347[/MATH], there are [MATH]5[/MATH] digits in this number therefore we write [MATH]4.[/MATH] and add [MATH]2 units[/MATH] to it's first digit which is [MATH]6[/MATH] and the answer is [MATH]log_{10}65347≈4.8[/MATH]
[MATH]C)[/MATH] If the first digit of our number is [MATH]8[/MATH] or [MATH]9[/MATH], we do the same as above but this time we put [MATH]9[/MATH] after the decimal point.
Example:
[MATH]log_{10}8546732[/MATH], there are [MATH]7[/MATH] digits in this number therefore we write [MATH]6.[/MATH] and put [MATH]9[/MATH] after the decimal point and the answer is [MATH]log_{10}8546732≈6.9[/MATH]
Now for [MATH]log_{10}400[/MATH] there are 3 digits in [MATH]400[/MATH] therefore we subtract one unit and because the first digit is from [MATH]2[/MATH] to [MATH]7[/MATH] we have to add [MATH]2units[/MATH] to it's first digit and the answer is [MATH]log_{10}400≈2.6[/MATH]thus the answer to your problem is [MATH]c>b>a[/MATH]

This is SUPER useful. What about numbers less than 1 (like log10 0.0004)? Are there similar general rules for converting square roots to decimals? What are the rules for natural log Ln?