||
Membaca
SMK Teknologi dan Rekayasa

Soal 7 Perpangkatan


Nilai\ x\ yang\ memenuhi\ persamaan\ 32^{2x-3}=\frac{1}{2}\ adalah\ ....\\a.\ \ -\frac{1}{5}\ \ \ \ \ \ b.\ \ \frac{2}{5}\ \ \ \ \ \ c.\ \ \frac{4}{5}\ \ \ \ \ \ d.\ \ \frac{7}{5}\ \ \ \ \ \ e.\ \ -\frac{5}{8}
Jawaban: d

\begin{array}{rcl}32^{2x-3}&=&\frac{1}{2}\\\\(2^{5})^{2x-3}&=&\frac{1}{2^{1}}\\\\2^{5(2x-3)}&=&2^{-1}\\\\5(2x-3)&=&-1\\\\10x-15&=&-1\\\\10x&=&14\\\\x&=&\frac{14}{10}\\\\x&=&\frac{7}{5}\end{array}

About Kalakay

Guru Matematika SMK

Diskusi

Belum ada komentar.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s

%d blogger menyukai ini: