Scientific Notation Application

[harpazo]

Section R.2
Algebra Essentials
Michael Sullivan
Textbook: College Algebra Edition 9

How long does it take a beam of light to reach Earth from the Sun when the Sun is 93,000,000 miles from Earth? Express your answer in scientific notation.

Solution:

Let x = speed that light travels through space is at 186,000 miles/second.

Let y = distance from Sun to Earth at 93,000,000 miles

Let T = length of time that it takes light to reach Earth from the Sun.

T = y/x

T = (93,000,000)/(186,000)

T = 500 seconds

It takes light 500 seconds to travel from Sun to Earth.

Express in scientific notation.

5.00 x 10^2

I believe this is right.

 

[pka]

That is correct.
I would have done \(\displaystyle \frac{93\cdot 10^6}{186\cdot 10^3}=\frac{10^3}{2}=5\cdot10^2\)

 

[harpazo]

That is correct.
I would have done \(\displaystyle \frac{93\cdot 10^6}{186\cdot 10^3}=\frac{10^3}{2}=5\cdot10^2\)

You divided one scientific notation by another scientific notation.

 

[JeffM]

You divided one scientific notation by another scientific notation.

Actually, if your measurements are in scientific notation, your calculations will also be in scientific notation and greatly simplified thereby.

[MATH]x = p * 10^m,\ y = q * 10^n, \ 1 \le p < 10,\ \text { and } 1 \le q < 10 \implies[/MATH]
[MATH]xy = pq * 10^{(m+n)} \text { and } 1 \le pq < 100 \implies[/MATH]
[MATH]xy = pq * 10^{(m+n)} \text { or } \dfrac{pq}{10} * 10^{(m+n+1)}.[/MATH]
Similarly,

[MATH]x = p * 10^m,\ y = q * 10^n, \ 1 \le p < 10,\ \text { and } 1 \le q < 10 \implies[/MATH]
[MATH]\dfrac{x}{y} = \dfrac{p}{q} * 10^{(m-n)}, \text { and } 0.1 < \dfrac{p}{q} \le 1 \implies[/MATH]
[MATH]\dfrac{x}{y} = \dfrac{p}{q} * 10^{(m-n)} \text { or } \dfrac{10p}{q} * 10^{(m-n-1)}.[/MATH]
When using a calculator, this is inconsequential, but when doing calculations by hand, it saves time and avoids errors.

 

[harpazo]

Actually, if your measurements are in scientific notation, your calculations will also be in scientific notation and greatly simplified thereby.

[MATH]x = p * 10^m,\ y = q * 10^n, \ 1 \le p < 10,\ \text { and } 1 \le q < 10 \implies[/MATH]
[MATH]xy = pq * 10^{(m+n)} \text { and } 1 \le pq < 100 \implies[/MATH]
[MATH]xy = pq * 10^{(m+n)} \text { or } \dfrac{pq}{10} * 10^{(m+n+1)}.[/MATH]
Similarly,

[MATH]x = p * 10^m,\ y = q * 10^n, \ 1 \le p < 10,\ \text { and } 1 \le q < 10 \implies[/MATH]
[MATH]\dfrac{x}{y} = \dfrac{p}{q} * 10^{(m-n)}, \text { and } 0.1 < \dfrac{p}{q} \le 1 \implies[/MATH]
[MATH]\dfrac{x}{y} = \dfrac{p}{q} * 10^{(m-n)} \text { or } \dfrac{10p}{q} * 10^{(m-n-1)}.[/MATH]
When using a calculator, this is inconsequential, but when doing calculations by hand, it saves time and avoids errors.

Your method of solving the problem is too advanced for me.

 

[Deleted member 4993]

Your method of solving the problem is too advanced for me.

I think you are just staring at the screen - that is why you probably have deemed it to be "too" advanced.

Take pencil and paper and follow the working step-by-step. At your level that practice will serve you well.

 

[JeffM]

I greatly doubt that. All I was saying is that if you multiply two numbers in scientific notation, you first multiply the two numbers that are not powers of 10 the normal way and calculate the powers of 10 by adding exponents.

[MATH](3 * 10^2)(2 * 10^3) = 6 * 10^5.[/MATH]
[MATH](3 * 10^2)(5 * 10^3) = 15 * 10^5.[/MATH]
Notice that the first is already in scientific notation, but the second is not.

[MATH]15 * 10^5 = \dfrac{15}{10} * 10 * 10^5 = 1.5 * 10^6.[/MATH]
In the second case, where the result is not quite in scientific notation, it is very easy to get it into proper form. And these are the only two cases there are.

There is nothing subtle about; it used to be taught to 9th graders when they were introduced to slide rules, a now obsolete calculating tool that ignored decimal points.

EDIT: As S. Khan suggested, try a few examples on your own, both multiplication and division. You will quickly see how simple it is. When I was a kid (before hand calculators), it was an essential computational trick. It is still useful for doing approximate calculations in your head (e.g., when negotiating). And it will convince you that scientific notation leads to simple calculations that involve multiplication, division, and exponentiation.

 

[pka]

You divided one scientific notation by another scientific notation.

I have absolutely no idea what your statement means.
I divided one number by another number just as you did. And got the same answer as you did.
In my case using exponents made the calculations easier.

 

[harpazo]

I greatly doubt that. All I was saying is that if you multiply two numbers in scientific notation, you first multiply the two numbers that are not powers of 10 the normal way and calculate the powers of 10 by adding exponents.

[MATH](3 * 10^2)(2 * 10^3) = 6 * 10^5.[/MATH]
[MATH](3 * 10^2)(5 * 10^3) = 15 * 10^5.[/MATH]
Notice that the first is already in scientific notation, but the second is not.

[MATH]15 * 10^5 = \dfrac{15}{10} * 10 * 10^5 = 1.5 * 10^6.[/MATH]
In the second case, where the result is not quite in scientific notation, it is very easy to get it into proper form. And these are the only two cases there are.

There is nothing subtle about; it used to be taught to 9th graders when they were introduced to slide rules, a now obsolete calculating tool that ignored decimal points.

EDIT: As S. Khan suggested, try a few examples on your own, both multiplication and division. You will quickly see how simple it is. When I was a kid (before hand calculators), it was an essential computational trick. It is still useful for doing approximate calculations in your head (e.g., when negotiating). And it will convince you that scientific notation leads to simple calculations that involve multiplication, division, and exponentiation.

I will try 10 more samples on my own.

 

[JeffM]

I will try 10 more samples on my own.

Good man.