Question

Cum se rezolvă un exercițiu de forma
8 la puterea n + 8 la puterea n+1= 18• 2 la puterea 2003?

Answer 1

Salutare!

$\bf 8^{n}+8^{n+1}=18 \cdot 2^{2003}$

$\bf 8^{n} \cdot(8^{n-n}+8^{n+1-n}) =9\cdot 2 \cdot 2^{2003}$

$\bf 8^{n} \cdot(8^{0}+8^{1}) =9\cdot 2^{2003 +1}$

$\bf 8^{n} \cdot(1+8) =9\cdot 2^{2004}$

$\bf 8^{n} \cdot 9 =9\cdot 2^{2004}\:\:\:|:9\:\:\text{(\it impartim toata relatia cu 9)}$

$\bf 8^{n} = 2^{2004}$

$\bf (2^{3})^{n} = 2^{2004}$

$\bf 2^{3\cdot n} = 2^{2004}$

$\bf 2^{3n} = 2^{2004}$

$\bf 3n = 2004\:\:\:|:3\:\:\text{(\it impartim toata relatia cu 3)}$

$\boxed{\bf n = 668}$

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