Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.5k views
in Exponents by (30.2k points)
closed by

If \(\frac{{2}^{m+n}}{{2}^{n-m}}\) = 16,and a = 21/10, then \(\frac{{a}^{2m+n-p}}{(a^{m-2n+2p})^{-1}}\) =

A. 2 

B.\(\frac{1}4\)

C.9

D. \(\frac{1}8\)

1 Answer

+1 vote
by (28.9k points)
selected by
 
Best answer

\(\frac{{2}^{m+n}}{{2}^{n-m}}\) = 2 2m + n - n + m = 2422m = 242m = 4m = 2Also a = 21/10

\(\frac{{a}^{2m+n-p}}{(a^{m-2n+2p})^{-1}}\) a2m+n-p x am-2n+2p

= a2m + m + n - 2n - p + 2p = a3m - n + p

\(= ({2}^{\frac{1}{10}})^{3m-n+p}\)

\(= ({2}^{\frac{1}{10}})^{3(2)-n+p}\)

\(= ({2}^{\frac{1}{10}})^{6-n+p}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...