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CBSE Class 8 Answered

Solve using laws of exponent a-3×b-7/a2×b-3    
Asked by garmaan204 | 03 Jun, 2020, 07:52: AM
answered-by-expert Expert Answer
begin mathsize 16px style fraction numerator straight a to the power of negative 3 end exponent cross times straight b to the power of negative 7 end exponent over denominator straight a squared cross times straight b to the power of negative 3 end exponent end fraction equals straight a to the power of negative 3 end exponent over straight a squared cross times straight b to the power of negative 7 end exponent over straight b to the power of negative 3 end exponent equals straight a to the power of negative 3 minus 2 end exponent cross times straight b to the power of negative 7 plus 3 end exponent equals straight a to the power of negative 5 end exponent cross times straight b to the power of negative 4 end exponent equals 1 over straight a to the power of 5 cross times 1 over straight b to the power of 4 end style
Answered by Yasmeen Khan | 03 Jun, 2020, 09:12: AM

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Solve using laws of exponent a-3×b-7/a2×b-3    
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