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Ahmad Ahmad

Bilangan Pangkat dan Akar

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ruas kiri di kali 1


karena 1= (4-3)


pakai identitas (a+b)(a-b)=(a^2-b^2)


sehingga diperoleh (4^64-3^64)=(4^x-3^y)


Maka, x-y=64-64=0.


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Berdasarkan bentuk $a^2 - b^2 = (a+b)(a - b)$, didapatkan

$\begin{align*} a^2 - b^2 &=(a+b)(a-b) \\ \therefore a + b &=\frac{a^2 - b^2}{a - b} \end{align*}$

Maka diperoleh 

$\begin{align*} 4^x - 3^y &=(3+4)(3^2+4^2)(3^4+4^4)(3^8+4^8)(3^{16}+4^{16})(3^{32}+4^{32}) \\ 4^x - 3^y & = \frac{3^2 - 4^2}{3 - 4} \cdot \frac{3^4 - 4^4}{3^2 - 4^2} \cdot \frac{3^8 - 4^8}{3^4 - 4^4} \cdot \frac{3^{16} - 4^{16}}{3^8 - 4^8} \cdot \frac{3^{32} - 4^{32}}{3^{16} - 4^{16}} \cdot \frac{3^{64} - 4^{64}}{3^{32} - 4^{32}} \\4^x - 3^y & = \frac{3^{64} - 4^{64}}{3 - 4} \\4^x - 3^y & = \frac{3^{64} - 4^{64}}{-1} \\4^x - 3^y & = 4^{64} - 3^{64} \\ \therefore x = y = 64 \end{align*}$

Maka $x-y=64-64=0$.

Edited by Wildan Bagus W

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