Persamaan Eksponen Bagian 4

Perasamaan Eksponen Bagian 4

# Persamaan Eksponen Berbentuk ${\color{Red} a^{f\left ( x \right )}=b^{g\left ( x \right )}}$

$\begin{array}{lcl} \textup{Jika}\; a^{f\left ( x \right )}=b^{g\left ( x \right )},\; \textup{dengan}\; a> 0,\; a\neq 1,\; b> 0,\; b\neq 1,\; \\\textup{maka}\; {\color{Red} f\left ( x \right )log\: a=g\left ( x \right )log\: b} \end{array}$

Karena penyelesaiannya dalam bentuk logaritma, berikut ini sifat-sifat logaritma:

$\begin{array}{lcl} 1.\; log\; a\cdot b=log\; a+log\; b\\ 2.\; log\; \frac{a}{b}=log\; a-log\; b\\ 3.\; log\; a^{n}=n\cdot log\; a\\ 4.\; a\: ^{a}\textrm{log }\; b=b\\ 5.\; ^{a}\textrm{log}\; b=\frac{log\; b}{log\; a}=\frac{1}{^{b}\textrm{log\: a}}\\ 6.\; ^{a^{m}}\textrm{log}\; b^{n}=\; ^{a}\textrm{log}\; b^{\frac{n}{m}}=\frac{n}{m}\cdot ^{a}\textrm{log}\; b\\ 7.\; ^{a}\textrm{log }\;b\; \cdot \; ^{b}\textrm{log}\; c=\; ^{a}\textrm{log}\; c \end{array}$

Soal dan Pembahasan

UMPTN 2001 Rayon C

$\begin{array}{lcl} 1.\; \textup{Nilai}\; x\; \textup{yang memenuhi}\; 8^{x+1}=24^{x-1}\; \textup{adalah .....}\\ \; \; \; \: \textup{A}.\; 1+6\; ^{2}\textrm{log}\: 3\\ \; \; \; \: \textup{B}.\; 1+4\; ^{2}\textrm{log}\: 3\\ \; \; \; \: \textup{C}.\; 1+6\; ^{3}\textrm{log}\: 2\\ \; \; \; \: \textup{D}.\; 1+4\; ^{3}\textrm{log}\: 2\\ \; \; \; \: \textup{E}.\; 1+6\; ^{5}\textrm{log}\: 2\\ \end{array}$

Pembahasan:

$\begin{array}{lcl} \: 8^{x+1}&=&24^{x-1}\\ 8^{x+1}&=&\left ( 8\cdot 3 \right )^{x-1}\\ 8^{x+1}&=&8^{x-1}\cdot 3^{x-1}\\ \frac{8^{x+1}}{8^{x-1}}&=&3^{x-1}\\ 8^{x+1-\left ( x-1 \right )}&=&3^{x-1}\\ 8^{x-x+1+1}&=&3^{x-1}\\ 8^{2}&=&3^{x-1}\\ 2\: \log \: 8&=&\left ( x-1 \right )\: \log 3\\ \log \: 8^{2}&=&\left ( x-1 \right )\: \log 3\\ x-1&=&\frac{\log \: 8^{2}}{\log 3}\\ x-1&=&^{3}\textrm{log}\: 8^{2}\\ \end{array}$
$\begin{array}{lcl} \: x-1&=&^{3}\textrm{log}\: 8^{2}\\ x&=&1+\: ^{3}\textrm{log}\: \left (2^{3} \right )^{2}\\ &=&1+\: ^{3}\textrm{log}\: 2^{6}\\ &=&1+\: 6\; ^{3}\textrm{log}\: 2 \end{array}$

Jawaban: _________________________ (C)

UMPTN 1995 Rayon C
$\begin{array}{lcl} 2.\; \textup{Penyelesaian persamaan }\; 2\left ( 25 \right )^{x+1}+5^{x+2}-3=0\; \textup{adalah .....}\\ \; \; \;\: \textup{A}.\; 1-\: ^{2}\textrm{log}\: 5\\ \; \; \;\: \textup{B}.\;- 1-\: ^{5}\textrm{log}\: 2\\ \; \; \;\: \textup{C}.\; 1+\: ^{5}\textrm{log}\: 2\\ \; \; \;\: \textup{D}.\;- 1-\: 5\: \textrm{log}\: 2\\ \; \; \;\: \textup{D}.\;1+\: 5\: \textrm{log}\: 2\\ \end{array}$

Pembahasan:

$\begin{array}{lcl} 2\left ( 25 \right )^{x+1}+5^{x+2}-3&=&0\\ 2\left ( 5^{2} \right )^{x+1}+5^{x}\cdot 5^{2}-3&=&0\\ 2\left ( 5 \right )^{2x+2}+5^{x}\cdot 25-3&=&0\\ 2\cdot 5^{2x}\cdot 5^{2}+5^{x}\cdot 25-3&=&0\\ 2\left ( 5^{x} \right )^{2}\cdot 25+5^{x}\cdot 25-3&=&0\\ 50\left ( 5^{x} \right )^{2}+ 25\left ( 5^{x} \right )-3&=&0\\\\ \textup{Misal}\; y=5^{x}\\ 50y^{2}+25y-3&=&0\\ \left ( 5y+3 \right )\left ( 10y-1 \right )&=&0\\ y=-\frac{3}{5}\; atau\; y=\frac{1}{10}\\ \textup{Ambil nilainya yang positif} \end{array}$

$\begin{array}{lcl} \textup{Maka} \; y=\frac{1}{10}\\\\ 5^{x}=\frac{1}{10}\\ 5^{x}=10^{-1}\\ x\: \textrm{log}\: 5&=&-\textrm{log}\: 10\\ x&=&\frac{-\textrm{log}\: 10}{\textrm{log}\: 5}\\ &=&-\: ^{5}\textrm{log}\: 5\cdot 2\\ &=&-\: \left (^{5}\textrm{log}\: 5+ ^{5}\textrm{log}\:2 \right )\\ &=&-\: \left (1+ ^{5}\textrm{log}\:2 \right )\\ &=&-1-\: ^{5}\textrm{log}\:2 \end{array}$

Jawaban: _________________________ (B)

UMB 2014 Kode 573

$\begin{array}{lcl} 3.\; \textup{Solusi Persamaan }\; 5^{2x+1}=10^{2x-1}\; \textup{adalah .....}\\ \; \; \; \: \textup{A}.\; ^{2}\textrm{log}\; 25\\ \; \; \; \:\textup{B}.\; ^{2}\textrm{log}\; 50\\ \; \; \; \:\textup{C}.\; ^{4}\textrm{log}\; 25\\ \; \; \; \:\textup{D}.\; ^{4}\textrm{log}\; 50\\ \; \; \; \:\textup{E}.\; ^{4}\textrm{log}\; 4\\ \end{array}$

Pembahasan:

$\begin{array}{lcl} 5^{2x+1}&=&10^{2x-1}\\ 5^{2x-1}\cdot 5^{2}&=&\left (5\cdot 2 \right )^{2x-1}\\ {\color{red} 5^{2x-1}}\cdot 25&=&{\color{Red} 5^{2x-1}}\cdot 2^{2x-1}\\ 25&=&2^{2x-1}\\ \textrm{log}\; 25&=&\left (2x-1 \right )\: \textrm{log}\; 2\\ 2x-1&=&\frac{ \textrm{log}\; 25}{\textrm{log}\; 2}\\ 2x-1&=&^{2}\textrm{log}\; 25\\ 2x&=&^{2}\textrm{log}\; 25+1\\ 2x&=&^{2}\textrm{log}\; 25+^{2}\textrm{log}\; 2\\ 2x&=&^{2}\textrm{log}\; 25\cdot 2\\ 2x&=&^{2}\textrm{log}\; 50\\ x&=&\frac{1}{2}\cdot \; ^{2}\textrm{log}\; 50\\ &=& ^{2}\textrm{log}\; 50^{\frac{1}{2}}\\ &=& ^{2^{2}}\textrm{log}\; 50\\ &=& ^{4}\textrm{log}\; 50\\ \end{array}$

Jawaban: _________________________ (D)

$\begin{array}{lcl} 4.\; \textup{Tentukan Penyelesaian dari }\; 2^{3x-1}=3^{x+1} \end{array}$

Pembahasan:

$\begin{array}{lcl} 2^{3x-1}&=&3^{x+1}\\ \left (3x-1 \right )\textrm{log\; 2}&=&\left (x+1 \right )\textrm{log\; 3}\\ 3x\;\textrm{log\; 2}\; -\;\textrm{log\; 2} &=&x\; \textrm{log\; 3}\; +\; \textrm{log\; 3}\\ 3x\;\textrm{log\; 2}\; -\; x\; \textrm{log\; 3}&=&\textrm{log\; 3}\; +\; \textrm{log\; 2}\\ x\left (3\;\textrm{log\; 2}\; - \textrm{log\; 3} \right )&=&\textrm{log\; 3}\; \cdot 2\\ x\left (\;\textrm{log\; 2}^{3}\; - \textrm{log\; 3} \right )&=&\textrm{log\; 6}\\ x\left (\;\textrm{log\; 8}\; - \textrm{log\; 3} \right )&=&\textrm{log\; 6}\\ x\cdot \textrm{log}\; \frac{8}{3}&=&\textrm{log\; 6}\\ x&=&\frac{\textrm{log\; 6}}{\textrm{log}\; \frac{8}{3}}\\ &=&\; ^{\frac{8}{3}}\textrm{log}\; 6 \end{array}$

SIMAK UI 2013 Kode 332

$\begin {array}{lcl} 5.\; \textup{Jika}\; 2^{\left ( x+2 \right )}+4^{\left ( x+1 \right )}=48,\; \textup{nilai dari}\; \frac{1}{x+1}\; \textup{adalah .....}\\ \; \; \; \: \textup{A}.\; ^{6}\textrm{log}\; 2\\ \; \; \; \: \textup{B}.\; \frac{1}{4}\\ \; \; \; \: \textup{C}.\; ^{2}\textrm{log}\; 3\\ \; \; \; \: \textup{D}.\; ^{2}\textrm{log}\; 6\\ \; \; \; \: \textup{A}.\; 3\\ \end{array}$

Pembahasan:

$\begin{array}{lcl} 2^{\left ( x+2 \right )}+4^{\left ( x+1 \right )}&=&48\\ 2^{\left ( x+1 \right )+1}+\left (2^{2} \right )^{\left ( x+1 \right )}&=&48\\ 2^{\left ( x+1 \right )}\cdot 2^{1}+2^{\left ( 2x+2 \right )}-48&=&0\\ 2^{\left ( x+1 \right )}\cdot 2+\left (2^{\left ( x+ 1\right )} \right )^{2}-48&=&0\\ 2\left ( 2^{x+1} \right )+\left ( 2^{x+1} \right )^{2}-48&=&0\\ \left ( 2^{x+1} \right )^{2}+2\left ( 2^{x+1} \right )-48&=&0 \\\\\textup{Misal}\; y=2^{x+1}\\ \; \; \; \; \; \; \; \; \; y^{2}+2y-48&=&0\\ \; \; \; \; \; \; \; \; \;\left ( y+8 \righ t )\left ( y-6 \right )&=&0\\ \end{array}$

$\begin{array}{lcl} \; \; \; \; \; \; \; \; \;y=-8\; \textup{atau}\; y=6\; \; \; \; \; \left ( \textup{ambil yang positif} \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \;2^{x+1}=6\;\\ \left ( x+1 \right )\: \textrm{log}\; 2=\textrm{log}\; 6\; \\ \; \; \; \; \; \; \; \; \; \; \; x+1=\frac{\textrm{log}\; 6}{\textrm{log}\; 2}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=\; ^{2}\textrm{log}\; 6-1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\; ^{2}\textrm{log}\; 6-\; ^{2}\textrm{log}\; 2\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\; ^{2}\textrm{log}\; \frac{6}{2}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\; ^{2}\textrm{log}\; 3\\\\ \textup{Nilai}\; \; \; \; \; \; \frac{1}{x+1}=\frac{1}{\; ^{2}\textrm{log}\; 3+1}\\ \end{array}$

$\begin{array}{lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\frac{1}{\; ^{2}\textrm{log}\; 3\; +\; ^{2}\textrm{log}\; 2}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\frac{1}{\; ^{2}\textrm{log}\; 3\; \cdot 2}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\frac{1}{\; ^{2}\textrm{log}\; 6}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; =\; ^{6}\textrm{log}\; 2 \end{array}$

Jawaban: _________________________ (A)

USM STIS 2015

$\begin{array}{lcl} 6.\; \textup{Penyelesaian persamaan}\; 6^{3+2x}-15\left ( 6^{1+x} \right )+6=0\; \textup{adalah .....}\\ \; \; \; \: \textup{A}.\; 1-\; ^{2}\textrm{log}\; 6\\ \; \; \; \:\textup{B}.\; -1-\; ^{2}\textrm{log}\; 6\\ \; \; \; \:\textup{C}.\; 1+\; ^{2}\textrm{log}\; 6\\ \; \; \; \:\textup{D}.\; -1-\; ^{6}\textrm{log}\; 2\\ \; \; \; \: \textup{E}.\; 1-\; ^{6}\textrm{log}\; 2\\ \end{array}$

Pembahasan:

$\begin{array}{lcl} 6^{3+2x}-15\left ( 6^{1+x} \right )+6&=&0\\ 6^{3}\cdot 6^{2x}-15\cdot 6\cdot 6^{x}+6&=&0\\ 6^{3}\cdot \left (6^{x}\right)^{2}-15\cdot 6\cdot \left (6^{x} \right )+6&=&0\; \; \; \; \left ( \textup{kedua ruas dibagi 6} \right )\\ 6^{2}\cdot \left (6^{x}\right)^{2}-15\cdot \left (6^{x} \right )+1&=&0\\\\ \textup{Misal }\; y=6^{x}\\ \; \; \; \; \; 6^{2}y^{2}-15y+1=0\\ \; \; \; \; \; 36y^{2}-15y+1=0\\ \left ( 12y-1 \right )\left ( 3y-1 \right )=0 \end{array}$

$\begin{array}{lcl} \left ( 12y-1 \right )\left ( 3y-1 \right )=0\\ 12y-1=0\; \textup{atau}\; 3y-1=0\\ y=\frac{1}{12}\; \textup{atau}\; y=\frac{1}{3}\\\\ \textup{Untuk}\; y=\frac{1}{12}\\ \; \; \; \; \; \; \; \; \; 6^{x}=\frac{1}{12}\\ \; \; \; \; \; \; \; \; \; 6^{x}=12^{-1}\\ \; \; x\; \textrm{log}\; 6=-\: \textrm{log}\; 12\\ \; \; \; \; \; \; \; \; \; \; x=\frac{-\: \textrm{log}\; 12}{\textrm{log}\; 6}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\:^{6}\textrm{log}\; 12 \end{array}$

$\begin{array}{lcl} \; \; \; \; \; \; \; \; \; \; \; \; \; =-\:^{6}\textrm{log}\; 12 \\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\:\left (^{6}\textrm{log}\; 6\cdot 2 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\:\left (^{6}\textrm{log}\; 6\; + \; ^{6}\textrm{log}\; 2 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\:\left (1 + \; ^{6}\textrm{log}\; 2 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\:1 - \; ^{6}\textrm{log}\; 2 \end{array}$

$\begin{array}{lcl} \textup{Untuk} \; y=\frac{1}{3}\\ \; \; \; \; \; \; \; \; \; 6^{x}=\frac{1}{3}\\ \; \; \; \; \; \; \; \; \; 6^{x}=3^{-1}\\ \; \; x\; \textrm{log}\; 6=-\; \textrm{log}\; 3\\ \; \; \; \; \; \; \; \; \; \; x=\frac{-\; \textrm{log}\; 3}{\textrm{log}\; 6}\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\; ^{6}\textrm{log}\; 3\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\; \left ( ^{6}\textrm{log}\; \frac{6}{2} \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\; \left ( ^{6}\textrm{log}\; 6\; -\; ^{6}\textrm{log}\; 2 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-\; \left ( 1 -\; ^{6}\textrm{log}\; 2 \right )\\ \; \; \; \; \; \; \; \; \; \; \; \; \; =-1+\; ^{6}\textrm{log}\; 2 \\ \end{array}$

Jawaban: _________________________ (D)

M@th

Selasa, 22 Agustus 2017

Persamaan Eksponen Bagian 4

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