9th Math Exercise 3.2 Question 3 || class 9

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In the name of Allah, the most gracious, the most merciful. In this video we are going to solve questions of 9th class math, science group, exercise three point two. So, stay with us.
We are going to start part one, of question number 3, of 9th class Math, Science group, exercise three point two.
Question 3, is, Find the numbers whose common logarithms are: And part one is three point five six two one.
Let see its solution.
We will let.
x. is equal to three point five six two one.
Here, we are taking log on both sides.
We take log on both sides. It becomes log x. is equal to log three point five six two one.
If we move log at left side to right side this will be converted to anti. So, next we need to find anti log of three point five six two one.
Here, we take point five six two one. And look in the table given above. Take point fifty six, then look under two, the value is three six four eight, then look under one, value is one.
So, add three six four eight and one, becomes three six four nine. So we write x. is equal to three six four nine.
Now, since characteristic is three so.
We have to place point according to characteristic which is three. If we move point from first digit to three points towards left then it will be three six four nine point zero. Hence we write only three six four nine.
Hence, this is our answer.
Next we are going to start part two of question 3, and Question 3, is, Find the numbers whose common logarithms are: And part two is one bar point seven four two seven.
Let see its solution.
We will let.
x. is equal to one bar point seven four two seven.
Here, we are taking log on both sides.
We take log on both sides. It becomes log x. is equal to log of one bar seven four two seven.
If we move log at left side to right side, this will be converted to anti. So, next we need to find anti log of one bar pint seven four two seven.
Here, we take point seven four two seven. And look in the table given above. Take point seventy four, then look under two, the value is five five two one, then look under seven, value is nine.
So, add five five two one and nine, becomes five five three zero. So we write x. is equal to five five three zero.
Now, since characteristic is one bar so.
We have to place point according to characteristic which is one bar. If we move point from first digit to one point towards right then it will be zero point five five three zero.
Hence this is our answer.
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