Soal dan Pembahasan Eksponen

1. UM UGM 2005 Kode 821

  Jika $\sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}$, maka $\frac{1}{a}+\frac{1}{b}$ = ....
    A. 25
    B. 20
    C. 15
    D. 10
    E. 5

    Pembahasan:
    KonseR
    $\begin{array}{lcl} \bigstar \; \sqrt{\left ( a+b \right )+\sqrt{a\cdot b}}=\sqrt{a}+\sqrt{b}\\\\ \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \sqrt{\left ( 0,3 \right )+\sqrt{0,08}}=\sqrt{a}+\sqrt{b}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \: \sqrt{\left ( 0,3 \right )+\sqrt{4}\cdot \sqrt{0,02}}=\sqrt{a}+\sqrt{b}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \sqrt{\left ( 0,3 \right )+2\sqrt{0,02}}=\sqrt{a}+\sqrt{b}\\ \sqrt{\left ( 0,2+0,1 \right )+2\sqrt{\left (0,2 \right )\left ( 0,1 \right )}}=\sqrt{a}+\sqrt{b}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \sqrt{0,2}+\sqrt{0,1}=\sqrt{a}+\sqrt{b} \end{array}$ 
   $\begin{array}{lcl} \sqrt{a}&=&\sqrt{0,2}\\ a&=&0,2\\ &=&\frac{2}{10}\\ &=&\frac{1}{5}\\\\ \sqrt{b}&=&\sqrt{0,1}\\ b&=&0,1\\ &=&\frac{1}{10}\\\\ \end{array}$ 
   $\begin{array}{lcl} \frac{1}{a}+\frac{1}{b}&=&\frac{1}{\frac{1}{5}}+\frac{1}{\frac{1}{10}}\\ &=&5+10\\ &=&15 \end{array}$ 

   Jawaban ____________________________________ (C) 

2. SPMB 2006 Kode320
    Diketahui$\begin{array}{lcl} 4^{x}=25\;\; \; dan \; 5^{y}=\frac{1}{8} \end{array}$. Bila $y$ 
    nyatakan dalam $x$, diperoleh $y$ = ....
    A. $\begin{array}{lcl} -\frac{3}{x} \end{array}$ 
    B. $\begin{array}{lcl} -\frac{2}{x} \end{array}$ 
    C. $\begin{array}{lcl} -\frac{x}{3} \end{array}$ 
    D. $\begin{array}{lcl} -\frac{x}{2} \end{array}$ 
    E. $\begin{array}{lcl} -\frac{2x}{3} \end{array}$ 

    Pembahasan:
    $\begin{array}{lcl} 4^{x}&=&25\\ \left ( 2^{2} \right )^{x}&=&25\\ \left (2^{x} \right )^{2}&=&25\\ 2^{x}&=&5\\\\ 5^{y}&=&\frac{1}{8}\\ \left ( 2^{x} \right )^{y}&=&8^{-1}\\ 2^{xy}&=&\left ( 2^{3} \right )^{-1}\\ 2^{xy}&=&2^{-3}\\ xy&=&-3\\ y&=&-\frac{3}{x} \end{array}$ 

    Jawaban ____________________________________ (A)