Basics of Logarithms-class 9 Logarithms-Exercise 3.3-Question 2 3 4 5
Basics of Logarithms-class 9 Logarithms-Exercise 3.3-Question 2 3 4 5
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Transcript
00:00bismillah hir rahman nir raheem. Assalam o alaikum dear students and viewers. In today's
00:15video we will solve the question number 2 of logarithms exercise 3.3. Inshallah. So
00:24we have in question number 2 express log of x plus 2 log of x minus 2 log of x plus 1
00:32minus log parenthesis x square minus 1. As a single logarithm you have to convert the
00:40expression given in terms of logarithm into a single expression. Solution let's start
00:48by analyzing the given logarithm expressions. First look at the plus and negative sign.
00:55Indicate that there are two kinds of formulas when we are going to convert into a single
01:04logarithm. A product and a quotient. After that you see the multiplier or coefficient
01:11of the product.
01:41Now we have successfully applied one rule. Now we will see how to use the product rule.
01:56Log x plus log x into log of x. This was the plus sign. What we have done is we have
02:07plused it. So here if we take the negative common then log x plus 1 raised to the power
02:14of 2 plus log into x square minus 1. If this is plus then it will be converted into a product
02:21and if it is minus then it will be converted into a single logarithm. Now our third step
02:26is to apply the quotient laws of logarithm because this is a negative sign. You have
02:32seen that the positive sign is being converted into a product. So here we have taken the
02:37negative common then both of them became positive and that also became a product. But this
02:42negative sign will go to the denominator.
02:49Log x plus 1 raised to the power of 2 into log into x square minus 1. This is our answer.
02:57Now we will go to question number 3. Write the following in the form of a single logarithm.
03:06The first part is log 21 plus log 5. So again after analyzing this question we come to know
03:16that there is a positive sign between the two terms of the logarithm expression. If
03:20there is a positive sign then it means if we write it in the form of a single logarithm
03:24then what will be the plus? It will be converted into a product. That is it will be converted
03:30into a product.
03:33Now we have the second part. We have to solve it.
03:38Log 25 minus 2 log 3. So in the first step what we will do is we will convert this multiplied
03:45by 2 into 3. In the second step we are going to convert this negative sign as a divisor
03:56in the form of coefficient.
03:58Now take log 21.
04:04Something is wrong here. So I will correct this.
04:09Log 25 over log 3.
04:26There can be another solution for this.
04:31Take the scale of 25.
04:35I will tell you how to write 2 log 5 5.
05:01square minus 2 log of 3.
05:09So this 2 will become power with this.
05:13log 5 minus 2 log 3.
05:26I will write the negative portion in the denominator.
05:38This 2 will be cancelled and log 5 divided by log 3 will come.
05:50Now we have part 3.
05:552 log of x minus 3 log of y.
06:01So you convert this 2 into x.
06:07and convert the negative sign into division.
06:10and take the portion log y raised to power 3 in the denominator.
06:14This is our solution.
06:16Now we will solve part 4.
06:19In part 4 we have log 5 plus log 6 minus log 2.
06:25So when we have a product then the numerator will go to product and the negative sign will go to divisor.
06:35Log 2 will go to product.
06:39Now we will start question number 4.
06:42Calculate the following
06:45Log 3 is the base.
06:482 into log 2 is equal to 21.
06:51So we have to solve this base change formula.
06:57So we have the solution.
06:59We know that log bn into log ab is equal to log bn and log ba.
07:09If you see the expression on the right hand side.
07:15The base of log is same in both numerator and denominator.
07:22And n is the first element as it is in the numerator.
07:29And in the denominator the base of a is the number.
07:38Now we will solve this equation.
07:41We need to find the b.
07:43Here its value is b is equal to 2.
07:45Log 2 is 3 raised to power 4.
07:48If you multiply 81 by 3 then you will get 81.
07:58So what we have done is
08:00Log 2 is 3 raised to power 4
08:04And the value of a is 3.
08:09So we have multiplied it with the value of a.
08:13And we have taken the number of a.
08:15Using power rule.
08:17Now we have 4.
08:20Multiply it with the value of a.
08:22Log base 2 is 3.
08:26Divided by log base 2 is 3.
08:28Both of them cancel each other.
08:30We have 4 answer.
08:32Now we will solve part 2 of equation number 4.
08:52Part 2 is log base 5 is 3.
08:56Multiplied by log base 3 is 25.
09:02We will use the same expression as above.
09:06Equation number 4 is log b into log a b.
09:11Log b is the first element as it is in the numerator.
09:16And in the denominator the base of b is 3.
09:23We have taken 3.
09:26Now we will multiply it with the value of 5.
09:31So that we can easily solve it.
09:33If the base is same then we will cancel it.
09:375 raised to power 2.
09:395 raised to power 3.
09:415 raised to power 3.
09:432 raised to power 3.
09:455 raised to power 3.
09:472 raised to power 3.
09:495 raised to power 3.
09:51Both of them cancel each other.
09:53We have 2 answer.
09:57Now we will solve part 5 of equation number 5.
10:02If log 2 is equal to 0.3010.
10:07Log 3 is equal to 0.47.
10:10Let me add 0 here.
10:130.4771 log is equal to 0.6990.
10:17Then find the values of the following log 32.
10:21First we have to convert 32 into factors.
10:27When we convert 32 into factors.
10:30Log 32 is equal to log 2 raised to power 5.
10:34Applying power rule.
10:36We have 5 into log 2.
10:395 multiplied by 5 is equal to 0.3010.
10:43Now we have the value of log 2.
10:450.3010.
10:47We have substituted here.
10:495 multiplied by 5 is equal to 1.5050.
10:55Now we will solve part 2.
10:58Log 24 solution.
11:02Again we will determine the factor of 24.
11:07Log 24 is equal to 2 into 2 into 2.
11:102 into 3 is equal to log 2 raised to power 3.
11:14This 3 is multiplied by this one.
11:27Now it is clear that this 3 is the power of 2.
11:32This 3 is multiplied by this whole factor or element.
11:36So applying product rule.
11:38We will get log 2 raised to power 3 plus log 3.
11:433 into log 2 plus log 3.
11:46Putting the value of log 2 and log 3.
11:493 into 2 is equal to 0.3010 plus 0.4771.
11:543 multiplied by 3 is equal to 0.903 plus 0.
12:04Now we have the answer of 1.380.
12:15Now we will move towards part 3.
12:23Log square root 3 into 1 by 3.
12:28Now we will convert the square root x into this form.
12:33Log square root 3 into 1 by 3 is equal to log 3 into 1 by 3 raised to power 1 by 2.
12:42The value of square root x is 1 by 2.
12:44So we have substituted it here.
12:46First apply power rule.
12:48We have 1 by 2 into log 10 by 3 is equal to 3 into 10.
12:539 plus 1 is equal to 10 by 3.
12:56By splitting 10 into its factor.
13:05Log 2 into 5.
13:07Now apply product and quotient laws of logarithm on this part of the equation.
13:16Simultaneously.
13:18It means that we will apply product rule on the denominator and quotient rule on 3.
13:24Log 2 plus log 5 minus log 3.
13:27Log 2 is equal to 0.3010.
13:33Log 5 is equal to 0.6990.
13:36Log 3 is equal to 0.4771.
13:39After multiplying all these, we have the answer of 0.5229.
13:43We have divided it by 2.
13:45So we have the answer of 0.26145.
13:49Now we will solve question number 4.
13:56Part 5 of question number 5.
13:59Log 8 by 3 solution.
14:02Now we will determine the factor of 8 in the numerator.
14:092 raised to power 3 divided by 3.
14:12Apply quotient rule.
14:15Log 2 raised to power 3 minus log 3.
14:18Power rule.
14:203 becomes coefficient of this log 2 minus log 3.
14:24Putting the value of log 2 and log 3.
14:273 into 0.3010 minus 0.471.
14:310.903 minus 0.4771 is equal to 0.4259.
14:38This is the answer.
14:44Now we will solve the fifth part of question number 5.
14:52Log 30.
14:53We will determine the factor of 30.
14:55Log 2 into 3 into 5.
14:57All log numbers are multiplying each other.
15:03So we are going to apply the quotient rule.
15:05Log 2 plus log 3 plus log 5.
15:08Substituting the given values of the respective logs.
15:12We will get 0.301 plus 0.4771 plus 0.6990.
15:191.4771.
15:22Answer.
15:26So our exercise is completed here.
15:29If you have any question or query.
15:31About any part of the any question.
15:34You can ask me in the comment section.
15:36And if you want to add something.
15:40You can give your opinion in the comment section.
15:42Thanks for watching.
15:43Assalam-o-Alaikum.