10th Class Maths 1st Lesson Real Numbers InText Questions and Answers<\/h3>\n
Do this<\/span><\/p>\n Question 1. Question 2. ii) 96 and 72 iii) 300 and 550 iv) 1860 and 2015 Think & Discuss<\/span><\/p>\n Question 1. Question 2. Question 3. Do this<\/span><\/p>\n Question 1. Question 2. ii) 50, 60 iii) 37, 49 Try this<\/span><\/p>\n Question 1. Do this<\/span><\/p>\n Question 1. Question 2. ii) \\(\\frac{7}{25}\\) iii) \\(\\frac{51}{64}\\) iv) \\(\\frac{14}{25}\\) v) \\(\\frac{80}{100}\\) Question 3. ii) \\(\\frac{2}{7}\\) iii) \\(\\frac{5}{11}\\) iv) \\(\\frac{10}{13}\\) Do this<\/span><\/p>\n Question 1. Think and Discuss<\/span><\/p>\n Question 1. Question 2. Question 3. Question 4. a) \\(3^{\\log _{3} 8}\\) b) \\(5^{\\log _{5} 10}\\) Do this<\/span><\/p>\n Question 1. ii) 100 = 5b<\/sup> iii) \\(\\frac{1}{81}\\) = 3c<\/sup> iv) 100 = 10z<\/sup> v) \\(\\frac{1}{256}\\) = 4a<\/sup> Question 2. ii) 235 \u00d7 437 iii) 2437 \u00d7 3568 Question 3. ii) \\(\\frac{373}{275}\\) iii) \\(\\frac{4525}{3734}\\) iv) \\(\\frac{5055}{3303}\\) Question 4. ii) log5<\/sub> 850<\/sup> iii) log 523<\/sup> iv) log 1024 Try this<\/span><\/p>\n Question 1. ii) log5<\/sub>625 = y iii) log10<\/sub>10000 = z iv) \\(\\log _{7} \\frac{1}{343}\\) = -a Question 2. ii) Find the value of logc<\/sub> \u221ac. iii) Find the value of log10<\/sub>0.001 iv) Find the value of \\(\\log _{\\frac{2}{3}} \\frac{8}{27}\\) AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers InText Questions AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers InText Questions and Answers. 10th Class Maths 1st Lesson Real Numbers InText Questions and Answers Do this Question 1. Find q and r for the following pairs … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5253"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5253"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5253\/revisions"}],"predecessor-version":[{"id":40586,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5253\/revisions\/40586"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nFind q and r for the following pairs of positive integers a and b, satisfying a = bq + r. (Page No. 3)
\ni) a = 13, b = 3
\nAnswer:
\n13 = 3 \u00d7 4 + 1
\nhere q = 4 ; r = 1
\nii) a = 8, b = 80
\nAnswer:
\nTake a = 80, b = 8
\n80 = 8 \u00d7 10 + 0 here q = 10 ; r = 0
\niii) a = 125, b = 5
\nAnswer:
\n125 = 5 \u00d7 25 + 0
\nhere q = 25 ; r = 0
\niv) a = 132, b = 11
\nAnswer:
\n132 = 11 \u00d7 12 + 0
\nhere q = 12 ; r = 0<\/p>\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\")
<\/p>\n
\nFind the HCF of the following by using Euclid division lemma,
\ni) 50 and 70 (Page No. 4)
\nAnswer:
\nFor given two positive integers a > b;
\nthere exists unique pair of integers q and r satisfying a = bq + r; 0\u2264r<b.
\n\u2234 70 = 50 \u00d7 1 + 20
\nHere a = 70, b = 50, q = 1, r = 20.
\nNow consider 50, 20
\n50 = 20 \u00d7 2 + 10
\nHere a = 50, b = 20, q = 2, r = 10.
\nNow taking 20 and 10.
\n20 = 10 \u00d7 2 + 0
\nHere the remainder is zero.
\n\u2234 10 is the HCF of 70 and 50.<\/p>\n
\nAnswer:
\n96 = 72 \u00d7 1 + 24
\n72 = 24 \u00d7 3 + 0
\n\u2234 HCF = 24<\/p>\n
\nAnswer:
\n550 = 300 \u00d7 1 + 250
\n300 = 250 \u00d7 1 + 50
\n250 = 50 \u00d7 5 + 0
\n\u2234 HCF = 50<\/p>\n
\nAnswer:
\n2015 = 1860 \u00d7 1 + 155
\n1860 = 155 \u00d7 12 + 0
\n\u2234 HCF = 155<\/p>\n<\/p>\n
\nFrom the above questions in ‘DO THIS’, what is the nature of q and r? (Page No. 3)
\nAnswer:
\nGiven: a = bq + r
\nq > 0 and r lies in between 0 and b
\ni.e. q > 0 and 0 \u2264 r < b<\/p>\n
\nCan you find the HCF of 1.2 and 0.12? Justify your answer. (Page No. 4)
\nAnswer:
\nGiven: 1.2 and 0.12
\nwe have 1.2 = \\(\\frac{12}{10}\\) = \\(\\frac{120}{100}\\)
\n0.12 = \\(\\frac{12}{100}\\)
\nNow considering the numerators 12 and 120, their HCF is 12.
\n\u2234 HCF of 1.2 and 0.12 is \\(\\frac{12}{100}\\) = 0.12
\ni.e., if x is a factor of y then x is the HCF of x and y.<\/p>\n
\nIf r = 0, then what is the relationship between a, b and q in a = bq + r of Euclid divison lemma? (Page No. 6)
\nAnswer:
\nGiven: r = 0 in a = bq + r then a = bq
\ni.e., b divides a completely.
\ni.e., b is a factor of a.<\/p>\n<\/p>\n
\nExpress 2310 as a product of prime factors. Also see how your friends have factorized the number. Have they done it as you ? Verify your final product with your friend’s result. Try this for 3 or 4 more numbers. What do you conclude? (Page No. 7)
\nAnswer:
\nGiven: 2310
\n2310 = 2 \u00d7 1155
\n= 2 \u00d7 3 \u00d7 385
\n= 2 \u00d7 3 \u00d7 5 \u00d7 77
\n2310 = 2 \u00d7 3 \u00d7 5 \u00d7 7 \u00d7 11
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-1.png\")
\nWe notice that this prime factorization is unique.
\nAnd also notice that prime factorization of any number is unique i.e., every composite number can be expressed as a product of primes and this factorization is unique.
\nE.g: 144 = 2 \u00d7 72
\n= 2 \u00d7 2 \u00d7 36
\n= 2 \u00d7 2 \u00d7 2 \u00d7 18
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 9
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3
\n= 24 \u00d7 32
\n320 = 2 \u00d7 160
\n= 2 \u00d7 2 \u00d7 80
\n= 2 \u00d7 2 \u00d7 2 \u00d7 40
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 20
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 10
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 5
\n= 26 \u00d7 5
\n125 = 5 \u00d7 25
\n= 5 \u00d7 5 \u00d7 5
\n= 53<\/sup><\/p>\n
\nFind the HCF and LCM of the following given pairs of numbers by prime factorization, (Page No. 8)
\ni) 120, 90
\nAnswer:
\nWe have 120 = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 5
\n= 23<\/sup> \u00d7 3 \u00d7 5
\n90 = 2 \u00d7 3 \u00d7 3 \u00d7 5
\n= 2 \u00d7 32<\/sup> \u00d7 5
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-2.png\")
\n\u2234 HCF = 2 \u00d7 3 \u00d7 5 = 30
\nLCM = 2<sup>3<\/sup> \u00d7 3<sup>2<\/sup> \u00d7 5 = 360<\/p>\n
\nAnswer:
\nWe have
\n50 = 2 \u00d7 5 \u00d7 5 = 2 \u00d7 52<\/sup>
\n60 = 2 \u00d7 2 \u00d7 3 \u00d7 5 = 22<\/sup> \u00d7 3 \u00d7 5
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-3.png\")
\n\u2234 HCF = 2 \u00d7 5 = 10
\nLCM = 22<\/sup> \u00d7 3 \u00d7 52<\/sup> = 300<\/p>\n
\nAnswer:
\nWe have
\n37 = 1 \u00d7 37
\n49 = 7 \u00d7 7 = 72<\/sup>
\n\u2234 HCF = 1
\nLCM = 37 \u00d7 72<\/sup>
\nNote: H.C.F. of two relatively prime numbers is 1 and LCM is equal to product of the numbers.<\/p>\n<\/p>\n
\nShow that 3n<\/sup> \u00d7 4m<\/sup> cannot end with the digit 0 or 5 for any natural numbers ‘n’ and’m’. (Page No. 8)
\nAnswer:
\nGiven number is 34<\/sup> \u00d7 4m<\/sup>.
\nSo the prime factors to it are 3 and 2 only.
\nI: but if a number want to be end with zero it should have 2 and 5 as its prime factors, but the given hasn\u2019t \u20185\u2019 as its prime factor.
\nSo it cannot be end with zero.
\nII : now if a number went to be end with 5 it should have \u20185\u2019 as its one of prime factors. But given 3n<\/sup> \u00d7 4m<\/sup> do not have 5 as a factor.
\nSo it cannot be end with 5.
\nHence proved.<\/p>\n
\nWrite the following terminating decimals in the form of p\/q, q \u2260 0 and p, q are co-primes.
\ni) 15.265
\nii) 0.1255
\niii) 0.4
\niv) 23.34
\nv) 1215.8
\nWhat can you conclude about the denominators through this process? (Page No. 10)
\nAnswer:
\ni) 15.265
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-4.png\")
\nii) 0.1255
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-5.png\")
\niii) 0.4
\n0.4 = \\(\\frac{4}{10}\\) = \\(\\frac{2}{5}\\)
\niv) 23.34
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-6.png\")
\nv) 1215.8
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-7.png\")
\nTwo and five are the factors for the denominator.<\/p>\n<\/p>\n
\nWrite the following rational numbers in the form of p\/q, where q is of the form 2n<\/sup>.5m<\/sup> where n, m are non-negative integers and then write the numbers in their decimal form. (Page No. 11)
\ni) \\(\\frac{3}{4}\\)
\nii) \\(\\frac{7}{25}\\)
\niii) \\(\\frac{51}{64}\\)
\niv) \\(\\frac{14}{25}\\)
\nv) \\(\\frac{80}{100}\\)
\nAnswer:
\ni) \\(\\frac{3}{4}\\)
\n\\(\\frac{3}{4}\\) = \\(\\frac{3}{2 \\times 2}\\) = \\(\\frac{3}{2^{2}}\\)
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-8.png\")
\nDecimal form of \\(\\frac{3}{4}\\) = 0.75<\/p>\n
\n\\(\\frac{7}{25}\\) = \\(\\frac{7}{5 \\times 5}\\) = \\(\\frac{7}{5^{2}}\\)
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-9.png\")
\nDecimal form of \\(\\frac{7}{25}\\) = 0.28<\/p>\n
\n\\(\\frac{51}{64}\\) = \\(\\frac{51}{2^{6}}\\)
\n[\u2235 64 = 2 \u00d7 32
\n= 22<\/sup> \u00d7 16
\n= 23<\/sup> \u00d7 8
\n= 24<\/sup> \u00d7 4 = 25<\/sup> \u00d7 2 = 26<\/sup>]
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-10.png\")
\nDecimal form of \\(\\frac{51}{64}\\) = 0.796875<\/p>\n
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-11.png\")
<\/p>\n
\n\\(\\frac{80}{100}\\) = \\(\\frac{80}{2^{2} \\times 5^{2}}\\) = \\(\\frac{80}{10^{2}}\\) = 0.80<\/p>\n<\/p>\n
\nWrite the following rational numbers as decimal form and find out the block of repeating digits in the quotient. (Page No. 11)
\ni) \\(\\frac{1}{3}\\)
\nii) \\(\\frac{2}{7}\\)
\niii) \\(\\frac{5}{11}\\)
\niv) \\(\\frac{10}{13}\\)
\nAnswer:
\ni) \\(\\frac{1}{3}\\)
\n\\(\\frac{1}{3}\\) = 0.3333…. = \\(0 . \\overline{3}\\)
\nBlock of digits, repeating in the quotient = period = 3.<\/p>\n
\nDecimal form of \\(\\frac{2}{7}\\) = 0.285714….
\nRepeating part\/period = 285714
\n\u2234 \\(\\frac{2}{7}\\) = \\(0 . \\overline{285714}\\)<\/p>\n
\nPeriod = 45
\nDecimal form of \\(\\frac{5}{11}\\) = 0.454545.
\n= \\(0 . \\overline{45}\\)<\/p>\n
\nDecimal form of \\(\\frac{10}{13}\\) = 0.769230.
\n= \\(0 . \\overline{769230}\\)
\nPeriod = 769230<\/p>\n
\nVerify the statement proved above for p = 2, p = 5 and for a2<\/sup> = 1, 4, 9, 25, 36, 49, 64 and 81. (Page No. 14)
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-16.png\")
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-17.png\")
\nFrom the above we can conclude that if a prime number \u2018p\u2019 divides a2<\/sup>, then it also divides a.<\/p>\n<\/p>\n
\nWrite the nature of y, a and x in y = ax<\/sup>. Can you determine the value of x for a given y? Justify your answer. (Page No. 17)
\nAnswer:
\ny = ax<\/sup> here a \u2260 0
\nWe can determine the value of \u2018x\u2019 for a given y.
\nfor example y = 5, a = 2
\nWe cannot express y = ax<\/sup> for y = 5, a = 2 and for y = 7, a = 3, we cannot express seven (7) as a power of 3.<\/p>\n
\nYou know that 21<\/sup> = 2, 41<\/sup> = 4, 81<\/sup> = 8 and 101<\/sup> = 10. What do you notice about the values of log2<\/sub> 2, log4<\/sub> 4, log8<\/sub> 8 and log10<\/sub> 10? What can you generalise from this?\u00a0 (Page No. 18)
\nAnswer:
\nFrom the graph log2<\/sub> 2 = log4<\/sub> 4 = log8<\/sub> 8 = log10<\/sub> 10 = 1
\nWe conclude that loga<\/sub> a = 1 where a is a natural number.<\/p>\n
\nDoes log10<\/sub> 0 exist? (Page No. 18)
\nAnswer:
\nNo, log10<\/sub> 0 doesn’t exist, i.e ax<\/sup> \u2260 0 \u2200 a, x \u2208 N.<\/p>\n
\nWe know that, if 7 = 2x<\/sup> then x = log2<\/sub> 7. Then what is the value of \\(2^{\\log _{2} 7}\\)? Justify your answer. Generalise the above by taking some more examples for \\(\\mathbf{a}^{\\log _{\\mathrm{a}} \\mathbf{N}}\\). (Page No. 21)
\nAnswer:
\nWe know that if 7 = 2x<\/sup> then x = log2<\/sub> 7
\nWe want to find the value of \\(2^{\\log _{2} 7}\\);
\nNow put log2<\/sub> 7 = x in the given
\n\u2234 \\(2^{\\log _{2} 7}\\) = 2x<\/sup> = 7 (given)
\n\u2234 \\(2^{\\log _{2} 7}\\) = 7
\nThus \\(\\mathbf{a}^{\\log _{\\mathrm{a}} \\mathbf{N}}\\) = N<\/p>\n
\nAnswer:
\nIf x = \\(3^{\\log _{3} 8}\\) then
\nlog3<\/sub> x = log3<\/sub> 8
\n\u21d2 x = 8<\/p>\n
\nAnswer:
\nIf y = \\(5^{\\log _{5} 10}\\)
\nthen log5<\/sub> y = log5<\/sub> 10
\n\u21d2 y = 10<\/p>\n
\nWrite the powers to which the bases to be raised in the following.\u00a0 (Page No. 18)
\ni) 64 = 2x<\/sup>
\nAnswer:
\n64 = 2x<\/sup>
\nWe know that
\n64 = 2 \u00d7 32
\n= 2 \u00d7 2 \u00d7 16
\n= 2 \u00d7 2 \u00d7 2 \u00d7 8
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 4
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2
\n64 = 26<\/sup>
\n\u21d2 x = 6<\/p>\n<\/p>\n
\nAnswer:
\nHere also 100 cannot be written as any power of 5.
\ni.e., there exists no integer for b such that 5b<\/sup> = 100<\/p>\n
\nAnswer:
\nWe know that 81 = 3 x 27
\n= 3 \u00d7 3 \u00d7 9
\n= 3 \u00d7 3 \u00d7 3 \u00d7 3
\n= 34
\n\u2234 \\(\\frac{1}{81}\\) = 3-4\u00a0<\/sup>\u00a0 [\u2235 a-m<\/sup> = \\(\\frac{1}{\\mathrm{a}^{\\mathrm{m}}}\\)]
\n\u2234 c = – 4<\/p>\n
\nAnswer:
\n100 = 102<\/sup>
\nz = 2<\/p>\n
\nAnswer:
\nWe know that 256 = 4 \u00d7 64
\n= 4 \u00d7 4 \u00d7 16
\n= 4 \u00d7 4 \u00d7 4 \u00d7 4
\n\u2234 \\(\\frac{1}{256}\\) = 4-4<\/sup>
\n\u2234 a = – 4<\/p>\n
\nExpress the logarithms of the following into sum of the logarithms.\u00a0 \u00a0(Page No. 19)
\ni) 35 \u00d7 46
\nAnswer:
\nlog xy = log x + log y
\nlog10<\/sub>35 \u00d7 46 = log10<\/sub>35 + log10<\/sub>46<\/p>\n
\nAnswer:
\nlog10<\/sub>235 \u00d7 437 = log10<\/sub>235 + log10<\/sub>437\u00a0 \u00a0[\u2235 log xy = log x + log y]<\/p>\n
\nAnswer:
\nlog10<\/sub> 2437 \u00d7 3568 = log10<\/sub>2437 + log10<\/sub>3568\u00a0 \u00a0[\u2235 log xy = log x + log y]<\/p>\n
\nExpress the logarithms of the follow\u00acing into difference of the logarithms.\u00a0 \u00a0(Page No. 20)
\ni) \\(\\frac{23}{34}\\)
\nAnswer:
\nlog10<\/sub> = \\(\\frac{23}{34}\\) = log10<\/sub> 23 – log10<\/sub> 34
\n[\u2235 log \\(\\frac{x}{y}\\) = log x – log y]<\/p>\n
\nAnswer:
\nlog10<\/sub> = \\(\\frac{373}{275}\\) = log10<\/sub> 373 – log10<\/sub> 275
\n[\u2235 log \\(\\frac{x}{y}\\) = log x – log y]<\/p>\n
\nAnswer:
\nlog10<\/sub> = \\(\\frac{4525}{3734}\\) = log10<\/sub> 4525 – log10<\/sub> 3734
\n[\u2235 log \\(\\frac{x}{y}\\) = log x – log y]<\/p>\n
\nAnswer:
\nlog10<\/sub> = \\(\\frac{5055}{3303}\\) = log10<\/sub> 5055 – log10<\/sub> 3303
\n[\u2235 log \\(\\frac{x}{y}\\) = log x – log y]<\/p>\n<\/p>\n
\nBy using the formula loga<\/sub>xn<\/sup> = n loga<\/sub> x, convert the following.\u00a0 \u00a0(Page No. 21)
\ni) log2<\/sub> 725<\/sup>
\nAnswer:
\nlog2<\/sub> 725<\/sup> = 25 log2<\/sub> 7<\/p>\n
\nAnswer:
\nlog5<\/sub> 850<\/sup> = 50 log5<\/sub> 8 = 50 log5<\/sub> 23
\n= 3 \u00d7 50 log5<\/sub>2 = 150 log5<\/sub>2<\/p>\n
\nAnswer:
\nlog 523<\/sup> = 23 log 5<\/p>\n
\nAnswer:
\nlog 1024 = log 210<\/sup> [\u2235 1024 = 210<\/sup>]
\n= 10 log 2<\/p>\n
\nWrite the following relation in exponential form and find the values of respective variables.\u00a0 \u00a0(Page No. 18)
\ni) log2<\/sub>32 = x
\nAnswer:
\nlog2<\/sub>32 = x
\n\u21d2 log2<\/sub>25<\/sup> = x\u00a0 \u00a0 \u00a0[\u2235 32 = 25<\/sup>]
\n\u21d2 5 log2<\/sub>2 = x\u00a0 \u00a0 \u00a0[\u2235 log am<\/sup> = m log a]
\n\u21d2 5 \u00d7 1 = x\u00a0 \u00a0 \u00a0 [\u2235 loga<\/sub> a = 1]
\n\u2234 x = 5<\/p>\n
\nAnswer:
\nlog5<\/sub>625 = y
\n\u21d2 log5<\/sub>4 = y\u00a0 \u00a0 [\u2235 625 = 54<\/sup>]
\n\u21d2 4 log5<\/sub> 5 = y\u00a0 \u00a0 \u00a0[\u2235 log am<\/sup> = m log a]
\n\u21d2 4 \u00d7 1 = y\u00a0 \u00a0 \u00a0[\u2235 loga<\/sub> a = 1]
\n\u2234 y = 4<\/p>\n<\/p>\n
\nAnswer:
\nlog10<\/sub>10000 = z
\n=> log10<\/sub>104<\/sup> = z\u00a0 \u00a0 \u00a0[\u2235 10000 = 10 \u00d7 10 \u00d7 10 \u00d7 10 = 104<\/sup>]
\n=> 4 log10<\/sub>10 = z\u00a0 \u00a0 \u00a0[\u2235 log am<\/sup> = m log a]
\n=> 4 \u00d7 1 = z\u00a0 \u00a0 \u00a0[\u2235 loga<\/sub> a = 1]
\n\u2234 z = 4<\/p>\n
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-18.png\")
<\/p>\n
\ni) Find the value of log2<\/sub>32. (Page No. 21)
\nAnswer:
\nlog2<\/sub> 32 = log2<\/sub> 25<\/sup>
\n[\u2235 32 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 = 25<\/sup>]
\n= 5 log2<\/sub> 2 [\u2235 log am<\/sup> = m log a]
\n= 5 \u00d7 1 [\u2235 loga<\/sub> a = 1]
\n= 5<\/p>\n
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-19.png\")
\n= \\(\\frac{1}{2}\\) \u00d7 1 [\u2235 loga<\/sub> a = 1]
\n= \\(\\frac{1}{2}\\)<\/p>\n<\/p>\n
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-20.png\")
<\/p>\n
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.in\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Chapter-1-Real-Numbers-InText-Questions-21.png\")
<\/p>\n","protected":false},"excerpt":{"rendered":"