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Mathematics

Find the value of n, when:

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}

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a2n−3×(a2)n+1(a4)−3=(a3)3÷(a6)−3⇒a2n−3×(a)2(n+1)(a)4×(−3)=(a)3×3÷(a)6×(−3)⇒a2n−3×(a)2n+2(a)−12=a9÷a−18⇒a(2n−3)+(2n+2)(a)−12=a9−(−18)⇒a2n−3+2n+2(a)−12=a9+18⇒a4n−1(a)−12=a27⇒a(4n−1)−(−12)=a27⇒a4n−1+12=a27⇒a4n+11=a27⇒4n+11=27⇒4n=27−11⇒4n=16⇒n=164⇒n=4\dfrac{a^{2n-3}\times(a^2)^{n+1}}{(a^4)^{-3}} = (a^3)^3 ÷ (a^6)^{-3}\\[1em] \Rightarrow \dfrac{a^{2n-3}\times(a)^{2(n+1)}}{(a)^{4\times(-3)}} = (a)^{3\times3} ÷ (a)^{6\times(-3)}\\[1em] \Rightarrow \dfrac{a^{2n-3}\times(a)^{2n+2}}{(a)^{-12}} = a^9 ÷ a^{-18}\\[1em] \Rightarrow \dfrac{a^{(2n-3)+(2n+2)}}{(a)^{-12}} = a^{9-(-18)}\\[1em] \Rightarrow \dfrac{a^{2n-3+2n+2}}{(a)^{-12}} = a^{9+18}\\[1em] \Rightarrow \dfrac{a^{4n-1}}{(a)^{-12}} = a^{27}\\[1em] \Rightarrow a^{(4n-1)-(-12)} = a^{27}\\[1em] \Rightarrow a^{4n-1+12} = a^{27}\\[1em] \Rightarrow a^{4n+11} = a^{27}\\[1em] \Rightarrow 4n+11 = 27\\[1em] \Rightarrow 4n = 27-11\\[1em] \Rightarrow 4n = 16\\[1em] \Rightarrow n = \dfrac{16}{4}\\[1em] \Rightarrow n = 4

If 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}, then n = 4

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