Find the value of n, where n is an integer and 2n–5 × 62n–4 = 1124×2. - Mathematics

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Sum

Find the value of n, where n is an integer and 2n–5 × 62n–4 = `1/(12^4 xx 2)`.

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Solution

We have, 2n–5 × 62n–4 = `1/(12^4 xx 2)`

⇒ `2^n/2^5 xx 6^(2n)/6^4 = 1/(12^4 xx 2)`  ......`[∵ a^(m-n) = a^m/a^n]`

⇒ `(2^n xx 6^(2n))/(2^5 xx 6^4) = 1/((2 xx 6)^4 xx 2)`  ......[∵ 12 = 6 × 2]

⇒ 2n × (62)n = `(2^5 xx 6^4)/(2^4 xx 6^4 xx 2)`  ......[By cross-multiplication] [∵ amn = (am)n and (a × b)m = am × am+n]

⇒ 2n × 36n = `(2^5 xx 6^4)/(2^5 xx 6^4)`  ......[∵ am × an = am+n]

⇒ 2n × 36n = 1

⇒ (2 × 36)n = 1  ......[∵ am × bm = (ab)m]

⇒ (72)n = (72)0  ......[∵ a0 = 1]

∴ n = 0  ......[∵ If am = an ⇒ m = n]

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Chapter 11: Exponents and Powers - Exercise [Page 344]

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NCERT Exemplar Mathematics Class 7
Chapter 11 Exponents and Powers
Exercise | Q 73. | Page 344

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