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Ajouté : 2026-06-01

[00:00:15]Okay, good morning students.

[00:00:18]Today we do some question from class 10 math length.

[00:00:24]Our topic is indices page number 199 exercise nine solve question number three.

[00:00:33]It's number three.

[00:00:37]problem. Now it's also here a we have 3 ^ x + 3 + 1 by 3 ^ x - 28 = 0.

[00:01:03]Now we are going to solve this question.

[00:01:07]3 a or 3^ x + 3 ^ x into 3 ^ 3 breaking power + 1 by 3 ^ x - 28 = 0 or 3 ^ x into 3 cube 3 into 3 into 3 27 + 1 by 3 ^x - 28 = 0.

[00:01:44]Let 3 ^x be a then it becomes 27 into a a into 27 + 1 by a - 28 =0 or taking lcm over here a lcm 27 a into a 27 A² + 1 - 28 into A 28 A = 0 or 27 A² + 1 - 28 A = A into 0 0 or 27 A² - 28 A + 1 equal to zero. Now we have to factoriize this or 27 a² - 27 + 1 a + 1 = 0.

[00:03:04]Sum is 28 product is 27 1 are 27 or 27 a² - 27 a - 1 into a - 1 a + 1 = 0 or from the first period taking 27 a common remaining a -1 and the second period minus - 1 is common remaining a -1 this = z then a -1 is common remaining 27 a -1 this equal to zero it is factorized now now either first factor = z or second factor =0 so either a minus one =0 0 or a = 0 + 1 1.

[00:04:08]But the value of a is 3 ^ x. So 3 ^ x = 1 in terms of base 3 ^0 or x = 0.

[00:04:24]Or second factor 27 a -1 = 0 or 27 a = 1 or a = 1 by 27 or a = 1 by 27 is 3 cube or 3 ^ x = 3 ^ - 3 or x = -3.

[00:04:57]So the value of x is 0 or - 3. This is the required solution.

[00:05:16]Okay students, now next question.

[00:05:21]Question number B.

[00:05:39]Okay. In 3 B we have 2 ^ x + 3 + 1 by 2 ^ x - 9 = 0.

[00:05:57]Now solving it odd ^ x + 3 that is 2 ^ x into ^ 3 + 1 by 2 ^ x - 9 = 0 or 2 ^ x into 2 cube 2 into 2 into 2 that is 8 + 1 I to the power x - 9 = 0 then to the power x b a then a into 8 + 1 by a - 9 = 0 taking LCM sum is A 8 A into A 8 A² + 1 - 9 into A 9 A = 0 or 8 A² + 1 - 9 A = A into Z 0 cross multiplication or it is - 9 a + 1 = 0 writing in descending order of power degree 2 degree 1 degree 0. Now we have to factoriize this sum is 9 product is 8 1 are 8.

[00:07:45]So we can write a² - 8 + 1 a + 1 = 0 or 8 a² - 8 a - 1 a + 1 = z on the first period taking a common remaining a minus 1 and the second period taking min -1 common remaining a -1 is equal to zero or a -1 common remaining 8 a -1 = 0 either a -1 =0 or a = 1 What the value of a is 2 ^ x. So 2^ x = 1 in the base 2^0 is 1.

[00:08:55]Base is same. So equating power x = 0.

[00:09:00]Or second factor a -1 = 0 or a = 1 or a = 1 by 8 or a = 1 by 2 cq or a is 2 ^ x = 1 by 2 that is 2 ^ - 3 or x = - 3.

[00:09:32]So the value of x is 0 or - 3.

[00:09:46]The value of x is 0 or - 3.

[00:09:52]This is the required solution.

[00:09:58]Now we go to next question.

[00:10:01]Question number 3 C.

[00:10:21]Now on C we have 2 ^ x + 1 upon 16 + 16 open the^ x + 1 = 65 upon 8.

[00:10:50]Now your solution are 2 ^ x + 1 that is 2 ^ x into ^ 1 all by 16 + 16 upon ^ x + 1 ^ x into 2 ^ 1 = to 65 by 8 or 2 ^ 1 is 2. So 2^ x upon 2 8 are 16. So open upon 8 2 are 16 and 2 are 16. So x upon 8 + 8 upon 2^x = to 65 by 8.

[00:11:42]Now we can suppose the power x be a then we get a by 8 + 8 by a = 65 upon 8. Now taking lcm 8 a is lcm. So a into a a² + 8 ha 64 = 65 by 8. Now 88 cancel out.

[00:12:17]Now remaining part is a² + 64 = 65 into a cross moduction it becomes 65 a or a² - 65 a + 64 = 0. Now we have to factoriize this.

[00:12:42]Your s is 65 and product is 1 into 64 64 or a² - 64 + 1 a + 64 = 0 or a² - 64 a - 1 a + 64 4 equals to zero.

[00:13:15]Now on the first period a is common remaining a min - 64.

[00:13:22]On the second pair -1 is common remaining a - 64 this equ= to zero or a - 64 is common remaining a -1 this = to zero then either a - 64 =0 or a = 64 but The value of a is 2 ^ x.

[00:13:55]So 2 ^ x = 64 to ^ 6 is 64 or x = 6.

[00:14:11]Or second factor a -1 =0 or a = 1 or 2^x = 1 that is 2 ^0 or x = 0.

[00:14:27]So the value of x is 0 or 6.

[00:14:39]Now the next question D.

[00:15:00]Okay students, now in question number D we have 7 ^ x + 1 by 7 ^ x = 49 whole 1 by 49 Okay, that 7 ^ x b a then it becomes a + 1 by a = 49 1 by 49 9.

[00:15:55]Then taking LCM on the left hand side a LCM a into a a² + 1 = 49 whole 1 by 49 that is 49 into 49 + 1 all upon 49 it becomes 49 into 49 9 are 81 1 carry over 8 9 are 36 + 8 44 9 for the 36 6 carry over 3 4 are 16 + 3 19 So adding 1 0 3 + 1 2 loss 1 1 4 02.

[00:17:22]So 1 4 02 upon 49 or cross multiplication 49 into a² 49 a² 49 1 are 49 = 142 into a 1 142 a 49 a² - 142 a + 49 = zero. Now we have to factoriize this or 49 a² - 1 4 0 1 + 1 a + 49 = 0 or 49 a² - 1 41 a - 1 a + 49 = 0 on the First taking 49 common 49 a common remaining a - 49 from second pair taking min -1 common remaining a - 49 this equal to zero again a - 49 common remaining 49 a -1 = z now we can do either first factor = 0 or second factor = to zero either or either a - 49 =0 or a = 49.

[00:19:05]What the value of a is 7 ^ x. So 7x = 49 is 7² or x = 2 or second factor 49 a - 1 = 0 or 49 a = 1 or a = 1 by 49 or a = 1 by 7² or 7 ^x = 7 ^ -2 or x = -2.

[00:19:46]So the value of x is 2 or - 2.

[00:20:05]Now the next question 3 E.

[00:20:27]Now question number 3 a we have ^ x + cos 1 by ^ x = to whole 1 by 4.

[00:20:44]Now solving 8 or 2^ x + 1 by 2 ^x = to 16 + 1 17 by 4. We can change mix fraction into simple fraction like this a whole b / by c = to a * c + b whole of c. In this way we can convert mixed fraction into simple fraction.

[00:21:20]Let the power x be a then it becomes a + 1 by a = 17 by 4. Then taking LCM on the left hand side a is LCM a * a a² + 1 this = 17 by 4. Then cross multiplication 4 into a² + 1 it becomes 4 a² + 4 = 17 * a 17 a or 4 a² - 17 a + 4 = 0. Now we have to factoriize this.

[00:22:15]Your sum is 17 and product is 4 4 are 16.

[00:22:20]So 4 a² - 16 + 1 * of a + 4 = 0 or a² - 16 a - 1 a + 4 = 0.

[00:22:42]Then from the first period we can take f a common taking a common remaining a -4 on the second pair taking -1 common remaining a - 4 this equals to zero again we have a -4 common remaining 4 a -1 this = 0.

[00:23:12]Now either a - 4 =0 or a = 4.

[00:23:21]But the value of a is 2 ^ x. So 2 ^ x = 4 in the base 2. We can write 2² or x = 2.

[00:23:34]And second factor are f - 1 =0 or a = 1 or a = 1 by 4 or a = 1 by 4 is 2² or value of a ^ x ^ x = 2 ^ - 2 or x = -2.

[00:24:09]So the value of x is 2 or -2. This is the required solution.

[00:24:24]Now next question on.

[00:24:45]Okay. On question number f we here 3 ^x + 1 by 3 ^x = 9 whole 1 by 9.

[00:25:03]Now solving 8 or 3 ^x + 1 by 3 ^x = 9 9 are 81 + 1 82 upon 9. Then suppose let 3 ^ x b a then it becomes a + 1 by a = 82 by 9 or taking lcm a lm a into a a² + 1 this = 82 by 9. Now cross multiplication 9 into a² + 1 it becomes 9 a² + 9 = 82 into a 82 a or 9 a² - 82 a + 9 = 0. Now we need to factoriize this. Your s is 82 and product is 9 9 are 81 or 9 a² - 81 + 1 * of a + 9 = 0 or 9 a² - 81 a - 1 a + 9 = 0.

[00:26:48]Or on the first pair taking 9 a common remaining A - 9 on the second pair taking -1 common remaining a - 9 = 0.

[00:27:04]Again a - 9 common remaining 9 a -1 = 0.

[00:27:11]either a - 9 =0 or a = 9.

[00:27:21]But the value of a is 3 ^ x. So 3 ^ x = 3² or x = 2. Similarly second factor are 9 a - 1 =0 or 9 a = 1 or a = 1 by 9 or a = 1 by 9 is 3² but a is 3 ^x so 3 ^x = 3 ^ -2 or x = - 2. So the value of x is 2 r - 2. This is the required solution.

[00:28:22]Now students next question 3 Z.

[00:28:43]Now in 3G the question is 5 ^ x + 1 by 5 ^ x = 25 whole 1 by 25 or 5 ^x + 1 by 5 ^ X = 25 25 + 625 + 1 626 upon 25 then we can suppose that ^ x b a then it becomes a + 1 by a = 626 upon 25.

[00:29:42]Then taking LCM or a into a a² + 1 all 1 upon a = 626 upon 25.

[00:29:58]Now cross multiplication 25 into a² + 1 25 a² + 25 = 626 into a 626 a or 25 a² - 626 a + 25 = Yeah.

[00:30:33]Now we have to factoriize this. Your sum is 626.

[00:30:38]Prog is 25 25 are 625 or 25 a² - 625 + 1 a + 25 = 0 or 25 a² - 625 a - 1 a + 25 = 0. On the first pair taking 25 a common remaining a minus 25.

[00:31:21]Now on the second pair taking minus1 common remaining a minus 25 = z.

[00:31:28]Again we have a - 25 is common remaining 25 a -1 = 0.

[00:31:39]Either a - 25 =0 or a = 25.

[00:31:49]But the value of a is 5 ^x. So 5^ x = 25 that is 5² in terms of base 5. So x = 2.

[00:32:03]Our second factor 25 a - 1 = 0 or 25 a = 1 or a = 1 by 25 or a = 1 by 5² or 5 x = 5 ^ -2 or x = -2.

[00:32:32]So the value of x is 2 or minus 2. This is the required solution.

[00:32:46]Now students we go to next question three. Yes.

[00:33:09]Now take we have 6 ^ x + 1 by 6 ^ x = 37 by 6.

[00:33:30]Now solving 8 we can suppose that 6 ^ x b a then it becomes a + 1 by a = 37 by 6.

[00:33:51]Taking LCM it becomes a into a a² + 1 all by a = 37 by 6. Then cross multiplication 6 into a² + 1 it makes 6 a² + 6 = to 37 into a it makes 37 a or 6 a² - 37 a + 6 = 0. Now we have to factoriize it. Sum is 37. Product is 6 into 6 36.

[00:34:27]So we can write 6 a² - 36 + 1 * of a + 6 = 0 or 6 a² - 36 a - 1 a + 6 = 0. On the first pair we can take 6 a common Remaining a - 6 on the second pair taking -1 common remaining a - 6 = 0 or again a - 6 common remaining 6 a -1 = 0.

[00:35:10]either first factor a - 6 = 0 or a = 6 but the value of a is 6 ^ x. So 6 ^ x = 6 is 6 ^ 1 or x = 1 or second factor 6 a - 1 =0 or 6 a = 1 or a = 1 by 6 or a is 6 ^ x = 1 by 6 that is 6 ^ -1 or x = -1. So the value of x is 1 or minus one. This is the required solution.

[00:36:20]Now, next question. Three.

[00:36:25]I now on 3 I the question is 10 ^ x + 1 upon 10 ^ x = and 1 upon 10.

[00:36:59]Now solving 8 let 10 ^ x be a then it becomes a + 1 by a = 101 upon a. Technical LCM we get a² + 1 all upon a = 101 upon 10 or cross multiplication 10 into a² + 1.

[00:37:34]It makes 10 a² + 10 = 101 into a 101 a² - 101 a + 10 = 0. Your sum is 101.

[00:37:54]Product is 10 10 are 100. So we can factoriize this 10² - 100 + 1 * of a + 10 =0 or 10 a² - 100 a - 1 a + 10 = z. On the first pair we can take 10 a common remaining a min - 10. On the second pair taking min -1 common remaining a minus 10 this equals to zero or we have a - 10 common remaining 10 a -1 this = z either a - 10 =0 or a = 10 but the value of 10 is value of a is 10 ^x. So 10 ^x = 10 is 10 ^ 1 or x = 1 or second factor 10 a - 1 =0 or 10 a = 1 or a = 1 by 10 or a is 10 ^ x = 1 by 10 is 10 ^ -1 or x = -1.

[00:39:46]So the value of x is -1 or 1. This is the required solution.

[00:40:01]Now we go to next question 3 Z 3 J we Have 3 ^ 2x + 1 all upon 3 ^ x = 82 upon 9.

[00:40:44]Now solving a r or 3 ^ 2x that is 3 ^ x² upon + 1 all upon 3 ^ x 2 into x 3x x into 3x = 82 by 9.

[00:41:11]Using power law we know that x ^ m power n = x ^ m into n mn by using power law of indices.

[00:41:23]Then we can suppose let 3 ^x be a then it becomes 3 ^ x² that is a² + 1 upon 3 ^ x that is a = 82 upon 9.

[00:41:42]Then cross multiplication we get 9 into a² + 1. It makes 9 a² + 9 = 82 into a 82 a or 9 a² - 82 a + 9 = 0. Now we have to factoriize this sum is 82 product is 9 9 are 81 or 9 a² - 81 + 1 * of a + 9 =0 or 9 a² - 81 a - 1 a + 9 = 0 on the first period taking 9 a common remaining a - 9 from The second we are taking -1 common remaining a - 9 =0.

[00:42:38]Again we have a - 9 common remaining 9 a -1 = 0. Now either or either first factor a - 9 =0 or a = 9 or a is 3 ^x. So 3 ^ x = 9 3² or x = 2.

[00:43:09]Our second factor 9 a - 1 = 0 or 9 a = 1 or a = 1 by 9 or a = 1 by 3² or 3 ^ x = 3 ^ -2 or x = -2 2.

[00:43:40]So the value of x is 2 or - 2. This is the required solution.

[00:43:54]Now next question.

[00:43:58]Okay.

[00:44:17]Now on k the question is 5 ^ 2x + 1 all upon 5 ^ x = 26 y by 5.

[00:44:40]Now solving 8 or 5 ^ 2x that is 5 ^ x whole² x into 2 2x 2 into x 3x + 1 by 5 ^x = 26 upon 5.

[00:45:02]Now suppose let 5 ^x be a then it becomes a² + 1 all by a = 26 by 5. Then cross multiplication a² + 1 into 5. It makes 5 a² + 5 = 26 into a 26 a or 5 a² - 26 a + 5 = 0. Now we have to factoriize this.

[00:45:45]Your sum is 26.

[00:45:48]The product is 5 into 5 25.

[00:45:52]So we can factoriize this.

[00:45:55]I² - 25 + 1 * of A + 5 = 0.

[00:46:06]Or 5 A² - 25 A - 1 A + 5 = 0.

[00:46:15]Or from the first pair taking 5 a common remaining a min - 5 and the second pair taking min -1 common remaining a - 5 =0 again a - 5 is common remaining 5 a -1 = 0. Now we can do either or either first factor equal to zero or second factor equal to zero.

[00:46:42]So either first factor a - 5 =0 or a = 5 or value of a is 5 ^x. So i ^ x = 5 ^ 1 or x = 1.

[00:47:05]Or second factor I a - 1 =0 or 5 a = 1 or a = 1 by 5 or 5 ^ x = 1x 5 that is 5 ^ -1 or x = -1.

[00:47:31]So the value of x is 1 or -1.

[00:47:40]This is the required solution.

[00:47:45]Now next question.

[00:48:01]on 3 L we have ^ 2x + 1 all 1 upon ^ x = 5 by 2.

[00:48:22]Now solving a r^ 2x that is 2 ^ x² x into 2x + 1 all by 2 ^ x = 5 upon 2. And you can suppose that e to the power x b a then it becomes a² + 1 all by a = 5 by 2. Then cross multiplication 2 into a² + 1. It makes a² + 2 = 5 into a it makes 5 a or a² - 5 a + 2 = 0. Now we have to factoriize this. Your sum is 5 and product is 2 to the 4.

[00:49:32]Or a² - 4 + 1 * of A + 2 = 0 or 2 A² - 4 A - 1 A + 2 = Z. And the first pair taking 2 a common remaining a -2 and the second pair taking -1 common remaining a - 2 = 0 or a - 2 common remaining a -1 = 0.

[00:50:18]It is factorized. Now either first factor= 0 either first factor is a minus 2= 0 or a = 2 or but the value of a is^x.

[00:50:33]So the power x = 2 is the power 1 or x = 1. Then second factor are 2 a - 1 = 0 or 2 a = 1 or a = 1 by 2 or the value of a is 2^ x. 2^ x = 1 by 2 that is 2^ -1 or x = -1.

[00:51:12]So the value of x is 1 or -1.

[00:51:25]So the value of x is 1 r -1. This is the required solution.

[00:51:32]Okay students, we have completed question number three.

[00:51:40]Okay, this was for this live class. In the next live we do next question. If you like my class, you can subscribe my channel and share to your friends.

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