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SMK Teknologi dan Rekayasa

Soal 8 Perpangkatan


Nilai\ x\ yang\ memenuhi\ persamaan\ \sqrt[3]{8^{x+3}}=(\frac{1}{32})^{x-3}\ adalah\ ....\\a.\ \ -2\ \ \ \ \ \ b.\ \ -1\ \ \ \ \ \ c.\ \ 0\ \ \ \ \ \ d.\ \ 1\ \ \ \ \ \ e.\ \ 2
Jawaban: e

\begin{array}{rcl}\sqrt[3]{8^{x+3}}&=&(\frac{1}{32})^{x-3}\\\\8^{\frac{x+3}{3}}&=&(\frac{1}{2^{5}})^{x-3}\\\\(2^{3})^{\frac{x+3}{3}}&=&(2^{-5})^{x-3}\\\\2^{3\times \frac{x+3}{3}}&=&2^{-5(x-3)}\\\\2^{x+3}&=&2^{-5x+15}\\\\x+3&=&-5x+15\\\\x+5x&=&15-3\\\\6x&=&12\\\\x&=&2\end{array}

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2 thoughts on “Soal 8 Perpangkatan

  1. knp 2 pangkat 3
    berubah jadi 2x gmn cara menghitungnya pak?

    Posted by fahrul saputra | 10 Februari 2014, 10:09 AM

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