Assassinaasim22 (b1)y2=2 Find the distance between Clinton and Brisbane if they are 6in apart on a map with a scale of 1in 11mi SHOW UR DANG WORK LOL whoever subs my channel Sleepy Samar on yt will get points Determine the integer whose product with (1) is bMath, 1115, vishal48 If 2^x = 3^y=12^z prove 1/x1/y1/z =0 Question If 2x = 3y = 6z then Prove that 1/x 1/y 1/z=0 Solution 2x = 6 z Þ 2 = 6z/x (i) 3y = 6 z Þ 3 = 6z/y (ii) Multiplying (i
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2^x=3^y=12^z show that 1/x 1/y 1/z=0
2^x=3^y=12^z show that 1/x 1/y 1/z=0- The Questions and Answers of If 2^x=3^y=6^z show that 1/x1/y1/z=0? To ask any doubt in Math download Doubtnut https//googl/s0kUoeQuestion If 2^x = 3^y =12^z show that 1/z= 1/y2/x
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If 2^x=3^y=6^z, prove that 1/x1/y1/z=0 anirbandas is waiting for your help Add your answer and earn pointsTake matha^x = b ^y = c^z = k/math (say) Then , matha = k^{1/x}, b = k^{1/y}/math and mathc = k^{1/z}/math Also, mathb^2 = ac/math So, mathk^{2/y If 2x = 3y = 12z, show that 1/z = 1/y 2/x Solution We are given, => 2 x = 3 y = 12 z = k (say) So, we get, => 12 = k 1/z => 2 × 3 × 2 = k 1/z => 2 2 × 3 = k 1/z => (k 1/x) 2 × (k) 1/y
(k) 1/x × (k) 1/y = (k)1/z By further calculation (k) 1/x 1/y = (k)1/z We get 1/x 1/y = – 1/z ⇒ 1/x 1/y 1/z = 0 Therefore, it is proved 25 If 2 x = 3 y = 12 z, prove that x = 2yz/y – z Solution It is given that 2 x = 3 y = 12 z Consider 2 x = 3 y = 12 z = k Here 2 x = k where 2 = (k) 1/x 3 y = k where 3 = (k) 1/y 12 z = k where 12 = (k)1/zAre solved by group of students and teacher of Class 9, which is also the largest student community of Class 9 If the answer is not available please wait for a while and a community member will probably answer this soon2 = (k) 1/x 3 y = k We can write it as 3 = (k) 1/y 6z = k We can write it as 6 = (k)1/z So we get 2 × 3 = 6 (k) 1/x × (k) 1/y = (k)1/z By further calculation (k) 1/x 1/y = (k)1/z We get 1/x 1/y = – 1/z 1/x 1/y 1/z = 0 Therefore, it is proved 25 If 2 x = 3 y = 12 z, prove that x = 2yz/y – z Solution It is given
If 2 x = 3 y = 12 z, show that 1/z = 1/y 2/x Solution 2 x = 3 y = (2 × 3 × 2) z 2 x = 3 y = (2 2 × 3) z 2 x = 3 y = (2 2z × 3 z) 2 x = 3 y = 12 z = k 2 = k 1/x 3 = k 1/y 12 = k 1/z 12 = 2 × 3 × 2 12 = k 1/z = k 1/y × k 1/x × k 1/x k 1/z = k 2/x 1/y 1/z = 1/y 2/x Question 6 If 2 x = 3 y = 6 −z, show that 1/x 1/y 1/z = 0 Solution If 2^x=3^y=6^z show that 1/x 1/y 1/z=0 Get the answers you need, now!Math, 0310, dishita4 2^x=3^y=12^z prove that 1/x1/y1/z=0
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Rd Sharma Solutions for Class 9 Math Chapter 2 Exponents Of Real Numbers are provided here with simple stepbystep explanations These solutions for Exponents Of Real Numbers are extremely popular among Class 9 students for Math Exponents Of Real Numbers Solutions come handy for quickly completing your homework and preparing for examsNCERT Solutions for Class 9 Math Chapter 2 Exponents Of Real Numbers are provided here with simple stepbystep explanations These solutions for Exponents Of Real Numbers are extremely popular among Class 9 students for Math Exponents Of Real Numbers Solutions come handy for quickly completing your homework and preparing for examsK^(1/x)=2 K^(1/y)=3 K^(1/z)=(1/6) Multiply these 3 equations to get K^(1/x)(1/y)(1/z)=1 Hence we can conclude that either k=1 ( which is totally false) or power is equal to "0" The correct answer is (1/x)(1/y)(1/z)=0
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Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreRD Sharma solutions for Mathematics for Class 9 chapter 2 (Exponents of Real Numbers) include all questions with solution and detail explanation This will clear students doubts about any question and improve application skills while preparing for board exams The detailed, stepbystep solutions will help you understand the concepts better and clear your confusions, if anyMath, 1115, Simrankanojia If 2^x = 3^y=12^z prove 1/x1/y1/z =0
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If 2 x = 3 y = 12 z, show that 1/z = 1/y 2/x Solution 2 x = 3 y = (2 × 3 × 2) z 2 x = 3 y = (2 2 × 3) z 2 x = 3 y = (2 2z × 3 z) 2 x = 3 y = 12 z = k 2 = k 1/x 3 = k 1/y 12 = k 1/z 12 = 2 × 3 × 2 12 = k 1/z = k 1/y × k 1/x × k 1/x k 1/z = k 2/x 1/y 1/z = 1/y 2/x Question 6 If 2 x = 3 y = 6 −z, show that 1/x 1/y 1/z = 0 Solution 2 x = 3 y = 6 −z 2 x = k 2 = k 1/x 3 y = k 3 = k 1/y 6 −z = k k
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