Find the value of n, when:
12−5×122n+1=1213÷12712^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^712−5×122n+1=1213÷127
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12−5×122n+1=1213÷127⇒12(−5)+(2n+1)=1213−7⇒12−5+2n+1=126⇒12−4+2n=126⇒−4+2n=6⇒2n=6+4⇒2n=10⇒n=102⇒n=512^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^7\\[1em] \Rightarrow 12^{(-5)+(2n+1)} = 12^{13 - 7}\\[1em] \Rightarrow 12^{-5+2n+1} = 12^{6}\\[1em] \Rightarrow 12^{-4+2n} = 12^{6}\\[1em] \Rightarrow -4+2n = 6\\[1em] \Rightarrow 2n = 6+4\\[1em] \Rightarrow 2n = 10\\[1em] \Rightarrow n = \dfrac{10}{2}\\[1em] \Rightarrow n = 5\\[1em]12−5×122n+1=1213÷127⇒12(−5)+(2n+1)=1213−7⇒12−5+2n+1=126⇒12−4+2n=126⇒−4+2n=6⇒2n=6+4⇒2n=10⇒n=210⇒n=5
If 12−5×122n+1=1213÷12712^{-5} \times 12^{2n+1} = 12^{13} ÷ 12^712−5×122n+1=1213÷127 then n = 5.
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Prove that:
[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1[xbxa]ab1[xcxb]bc1[xaxc]ca1=1
11+xa−b+11+xb−a=1\dfrac{1}{1+x^{a-b}}+\dfrac{1}{1+x^{b-a}}=11+xa−b1+1+xb−a1=1
a2n−3×(a2)n+1(a4)−3=(a3)3÷(a6)−3\dfrac{a^{2n-3}\times(a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 ÷ (a^6)^{-3}(a4)−3a2n−3×(a2)n+1=(a3)3÷(a6)−3
Simplify:
a2n+3.a(2n+1)(n+2)(a3)2n+1.an(2n+1)\dfrac{a^{2n+3}.a^{(2n+1)(n+2)}}{(a^3)^{2n+1}.a^{n(2n+1)}}(a3)2n+1.an(2n+1)a2n+3.a(2n+1)(n+2)
