বৈজ্ঞানিক সংখ্যাপদ্ধতি: গুণ

(Video) Multiplication with Scientific Notation

by mathman1024

This video by mathman1024 covers multiplication with numbers written in scientific notation. Here's the formula for how it's done: $$(a\times10^m)\cdot(b\times10^n)=ab\times10^{m+n}\\ \text{where }1\leq ab<10$$

$\large{\text{Summary}}$

He begins with the example: $(2.3\times10^4)\cdot(3\times10^7)$. Because everything is being multiplied this can be rearranged as follows. The first step works because of the commutative property of multiplication. $$\begin{align} &(2.3\times10^4)\cdot(3\times10^7)\\ &=2.3\cdot3\cdot10^4\cdot10^7\\ &=6.9\times10^{4+7}\\ &=6.9\times10^{11} \end{align}$$

The last step is true by the product rule for exponents. We are simply adding powers together.

So all you need to do is multiply the decimal parts together and then add the exponents for the $10$'s parts. There's one final thing you have to be weary of when doing this and we will see it in the next example.

He looks at the problem, $(7.1\times10^{-10})\cdot(5.8\times10^3)$. $$\begin{align} &(7.1\times10^{-10})\cdot(5.8\times10^3)\\ &=7.1\cdot5.8\cdot10^{-10}\cdot10^3\\ &=41.18\times10^{-10+3}\\ &=41.18\times10^{-7} \end{align}$$ So we're done right? Not exactly. Look at the new decimal number we have, $41.18$. This needs to be between $1$ and $10$ in order to be in scientific notation. So let's move the decimal over to the left one position. $$41.18\rightarrow4.118$$ Now that's a number that works for us! But we moved the decimal place over one so we have to represent this new change in our exponent. But do we add $1$ to the exponent or subtract $1$? This is where it can get tricky. I can assure you that even I have trouble keeping track whether you increase or decrease. But he gives a good analogy to help you understand.

So $41.18$ is too big for scientific notation. Assume this is the same as having too much money in your wallet. In order to get rid of money you have to deposit some in the bank, $10^{-7}$. If we deposit some, our wallet has less money, $4.118$. But now that money has gone to the bank, so the money in the bank has increased, $10^{-7\color{#29A603}{+1}}=10^{-6}$.

Therefore, the answer to the problem is $$\boxed{4.118\times10^{-6}}$$

(Video) Multiply Numbers Written in Scientific Notation

by Mathispower4u

Mathispower4u has a great video that explains multiplication with numbers in Scientific Notation.

$\large{\text{Summary}}$

The main idea is to separate the $2$ parts of the number and multiply these together individually like below: $$(a\times 10^n)\cdot (b\times 10^m)=(a\cdot b\times 10^{n+m})$$

Let's look at the first example: $$(1.2\times10^4)(5.3\times10^3)$$ Because multiplication is commutative, we can multiply this in any order we want. Let's multiply the constant parts together, and then multiply the power parts together. $$\begin{align} &(1.2\times10^4)(5.3\times10^3)\\ &=(1.2\cdot5.3)\times(10^4\cdot10^3)\\ &=6.36\times(10^4\cdot10^3)\\ &=6.36\times10^{4+3}\\ &=6.36\times10^7\\ \end{align}$$

Example

One thing to be careful about is when the product of the constant terms gives a value outside of the range of $1$ through $10$. If this happens, you have to readjust the decimal at the end.

An example of this is: $$(9\times10^1)(7\times10^9)$$

Let's solve. $$\begin{align} &(9\times10^1)(7\times10^9)\\ &=(9\cdot7)\times(10^1\cdot10^9)\\ &=63\times(10^1\cdot10^9)\\ &=63\times10^{1+9}\\ &=63\times10^{10}\\ \end{align}$$ As you can see, $63$ is not between $1$ and $10$. Therefore, we need to move the decimal place to the left one space. This will increase the exponent by one as well. $$\begin{align} &63\times10^{10}\\ &=6.3\times10^1\times10^{10}\\ &=6.3\times10^{1+10}\\ &=6.3\times10^{11} \end{align}$$

Multiplying and Dividing Scientific Notation

Example One

Let's jump right in with a problem: $$(2\times10^3)\cdot(4\times10^{20})$$ The order in which you multiply doesn't matter. We start by multiplying the constants, and then the powers of $10$. $$(2\cdot4)\times(10^3\cdot10^{20})$$ When multiplying powers with the same base, you just add the exponents. $$\begin{align} &(2\cdot4)\times(10^3\cdot10^{20})\\ &=8\times(10^3\cdot10^{20})\\ &=8\times10^{3+20}\\ &=\boxed{8\times10^{23}} \end{align}$$

Example Two

Now, let's look at a trickier problem. $$(3.5\times10^{-2})\cdot(3\times10^5)$$ We use the same method as before but there will be one extra step. $$\begin{align} &(3.5\times10^{-2})\cdot(3\times10^5)\\ &=(3.5\cdot3)\times(10^{-2}\cdot10^5)\\ &=10.5\times(10^{-2}\cdot10^5)\\ &=10.5\times10^{-2+5}\\ &=10.5\times10^3 \end{align}$$

Now be careful, $10.5$ is not in scientific notation anymore because isn't between $1$ and $10$. We have to move the decimal place to the left one space. That ensures the decimal in our final answer is between $1$ and $10$.

$$\begin{align} &10.5\times10^3\\ &=1.05\times10^1\times10^3\\ &=1.05\times10^{1+3}\\ &=\boxed{1.05\times10^4} \end{align}$$