Exercise Zone : Logaritma [2]
Table of Contents
Berikut ini adalah kumpulan soal mengenai Logaritma. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.
Tipe:
No.
Sederhanakanlah\(\log10^{\frac12}+\log\sqrt{10^3}\)
ALTERNATIF PENYELESAIAN
\begin{aligned}
\log10^{\frac12}+\log\sqrt{10^3}&=\dfrac12+\log10^{\frac32}\\[3.7pt]
&=\dfrac12+\dfrac32\\[3.7pt]
&=\dfrac42\\
&=\boxed{\boxed{2}}
\end{aligned}
Jadi, \(\log10^{\frac12}+\log\sqrt{10^3}=2\).
No.
Sederhanakanlah\(^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}\)
ALTERNATIF PENYELESAIAN
\begin{aligned}
^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}&=^2\negthinspace\log2^{\frac12}:{^2\negthinspace\log\sqrt{2^3}}\\[3.7pt]
&=\dfrac12:{^2\negthinspace\log2^{\frac32}}\\[3.7pt]
&=\dfrac12:\dfrac32\\[3.7pt]
&=\dfrac12\times\dfrac23\\
&=\boxed{\boxed{\dfrac13}}
\end{aligned}
Jadi, \(^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}=\dfrac13\).
No.
SederhanakanlahALTERNATIF PENYELESAIAN
\begin{aligned}
^4\negthinspace\log2+{^4\negthinspace\log32}&={^{2^2}\negthinspace\log2}+{^{2^2}\negthinspace\log2^5}\\
&=\dfrac12+\dfrac52\\[3.7pt]
&=\dfrac62\\
&=\boxed{\boxed{3}}
\end{aligned}
Jadi, 4log 2 + 4log 32 = 3.
No.
JikaALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{3-3\log^2xy}{1-\log x^3y^2+2\log x\sqrt{y}}&=\dfrac{3\left(1-\log^2xy\right)}{1-\left(\log x^3y^2-\log\left(x\sqrt{y}\right)^2\right)}\\[3.7pt]
&=\dfrac{3\left(1+\log xy\right)\left(1-\log xy\right)}{1-\left(\log x^3y^2-\log x^2y\right)}\\[3.7pt]
&=\dfrac{3\left(1+\log xy\right)\left(1-\log xy\right)}{1-\log\dfrac{x^3y^2}{x^2y}}\\[3.7pt]
&=\dfrac{3\left(1+\log xy\right)\cancel{\left(1-\log xy\right)}}{\cancel{1-\log xy}}\\
&=\boxed{\boxed{3+3\log xy}}
\end{aligned}
Jadi, \(\dfrac{3-3\log^2xy}{1-\log x^3y^2+2\log x\sqrt{y}}=3+3\log xy\).
No.
Diketahui nilai dari- \(\dfrac{3-2(a-1)(b-1)}{a+(a-1)(b-1)}\)
- \(\dfrac{6-4(a-1)(b-1)}{(2a+3)(b-1)}\)
- \(\dfrac{3+2(a-1)(b-1)}{a-(a-1)(b-1)}\)
- \(\dfrac{6+4(a-1)(b-1)}{(2a-3)(b-1)}\)
- \(\dfrac{3+2(a-1)(b-1)}{(a-1)(b-1)-a}\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
^{45}\negthinspace\log72=a\\
\dfrac{^3\negthinspace\log72}{^3\negthinspace\log45}&=a\\
\dfrac{^3\negthinspace\log\left(3^2\cdot2^3\right)}{^3\negthinspace\log\left(3^2\cdot5\right)}&=a\\
\dfrac{2+3\ ^3\negthinspace\log2}{2+{^3\negthinspace\log5}}&=a\\
2+3\ ^3\negthinspace\log2&=2a+a\ ^3\negthinspace\log5\\
3\ ^3\negthinspace\log2-a\ ^3\negthinspace\log5&=2(a-1)
\end{aligned}\)
\(\begin{aligned} ^{20}\negthinspace\log180&=b\\ \dfrac{^3\negthinspace\log180}{^3\negthinspace\log20}&=b\\ \dfrac{^3\negthinspace\log\left(3^2\cdot2^2\cdot5\right)}{^3\negthinspace\log\left(2^2\cdot5\right)}&=b\\ \dfrac{2+2\ ^3\negthinspace\log2+{^3\negthinspace\log5}}{2\ ^3\negthinspace\log2+{^3\negthinspace\log5}}&=b\\ 2+2\ ^3\negthinspace\log2+{^3\negthinspace\log5}&=2b\ ^3\negthinspace\log2+b\ ^3\negthinspace\log5\\ 2(b-1)\ ^3\negthinspace\log2+(b-1)\ ^3\negthinspace\log5&=2 \end{aligned}\)
\(\begin{aligned} 2(b-1)\ ^3\negthinspace\log2+(b-1)\ ^3\negthinspace\log5&=2\qquad&\color{red}{\times3}\\ 3\ ^3\negthinspace\log2-a\ ^3\negthinspace\log5&=2(a-1)\qquad&\color{red}{\times2(b-1)} \end{aligned}\)
\(\begin{aligned} 6(b-1)\ ^3\negthinspace\log2+3(b-1)\ ^3\negthinspace\log5&=6\\ 6(b-1)\ ^3\negthinspace\log2-2a(b-1)\ ^3\negthinspace\log5&=4(a-1)(b-1)\qquad-\\\hline (3+2a)(b-1)\ ^3\negthinspace\log5&=6-4(a-1)(b-1)\\ ^3\negthinspace\log5&=\dfrac{6-4(a-1)(b-1)}{(3+2a)(b-1)}\\ &=\boxed{\boxed{\dfrac{6-4(a-1)(b-1)}{(2a+3)(b-1)}}} \end{aligned}\)
\(\begin{aligned} ^{20}\negthinspace\log180&=b\\ \dfrac{^3\negthinspace\log180}{^3\negthinspace\log20}&=b\\ \dfrac{^3\negthinspace\log\left(3^2\cdot2^2\cdot5\right)}{^3\negthinspace\log\left(2^2\cdot5\right)}&=b\\ \dfrac{2+2\ ^3\negthinspace\log2+{^3\negthinspace\log5}}{2\ ^3\negthinspace\log2+{^3\negthinspace\log5}}&=b\\ 2+2\ ^3\negthinspace\log2+{^3\negthinspace\log5}&=2b\ ^3\negthinspace\log2+b\ ^3\negthinspace\log5\\ 2(b-1)\ ^3\negthinspace\log2+(b-1)\ ^3\negthinspace\log5&=2 \end{aligned}\)
\(\begin{aligned} 2(b-1)\ ^3\negthinspace\log2+(b-1)\ ^3\negthinspace\log5&=2\qquad&\color{red}{\times3}\\ 3\ ^3\negthinspace\log2-a\ ^3\negthinspace\log5&=2(a-1)\qquad&\color{red}{\times2(b-1)} \end{aligned}\)
\(\begin{aligned} 6(b-1)\ ^3\negthinspace\log2+3(b-1)\ ^3\negthinspace\log5&=6\\ 6(b-1)\ ^3\negthinspace\log2-2a(b-1)\ ^3\negthinspace\log5&=4(a-1)(b-1)\qquad-\\\hline (3+2a)(b-1)\ ^3\negthinspace\log5&=6-4(a-1)(b-1)\\ ^3\negthinspace\log5&=\dfrac{6-4(a-1)(b-1)}{(3+2a)(b-1)}\\ &=\boxed{\boxed{\dfrac{6-4(a-1)(b-1)}{(2a+3)(b-1)}}} \end{aligned}\)
Jadi, \(^3\negthinspace\log5=\dfrac{6-4(a-1)(b-1)}{(2a+3)(b-1)}\).
JAWAB: B
JAWAB: B
No.
Hasil dari \(\left(\dfrac{^9\negthinspace\log4\cdot{^8\negthinspace\log3}+{^3\negthinspace\log9}}{^3\negthinspace\log54-{^3\negthinspace\log2}}\right)^2=\)- \(\dfrac{49}{81}\)
- \(\dfrac79\)
- \(\dfrac76\)
- \(\dfrac13\)
- \(\dfrac37\)
ALTERNATIF PENYELESAIAN
\begin{aligned}
\left(\dfrac{^9\negthinspace\log4\cdot{^8\negthinspace\log3}+{^3\negthinspace\log9}}{^3\negthinspace\log54-{^3\negthinspace\log2}}\right)^2&=\left(\dfrac{^{3^2}\negthinspace\log2^2\cdot{^{2^3}\negthinspace\log3}+2}{^3\negthinspace\log\dfrac{54}2}\right)^2\\
&=\left(\dfrac{\dfrac13\cdot{^3\negthinspace\log2}\cdot{^2\negthinspace\log3}+2}{^3\negthinspace\log27}\right)^2\\
&=\left(\dfrac{\dfrac13+2}3\right)^2\\
&=\left(\dfrac{\dfrac73}3\right)^2\\
&=\left(\dfrac79\right)^2\\
&=\boxed{\boxed{\dfrac{49}{81}}}
\end{aligned}
Jadi, \(\left(\dfrac{^9\negthinspace\log4\cdot{^8\negthinspace\log3}+{^3\negthinspace\log9}}{^3\negthinspace\log54-{^3\negthinspace\log2}}\right)^2=\dfrac{49}{81}\).
JAWAB: A
JAWAB: A
No.
Jika- \(\dfrac{2b+a}{1+a}\)
- \(\dfrac{2ab+a}{2+a}\)
- \(\dfrac{3ab+1}{1+a}\)
- \(\dfrac{3ab+a}{1+a}\)
- \(\dfrac{3b+1}{1+a}\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
3^a&=5\\
^3\negthinspace\log5&=a
\end{aligned}\)
\(\begin{aligned} 5^b&=2\\ ^5\negthinspace\log2&=b \end{aligned}\)
\(\begin{aligned} ^3\negthinspace\log5\cdot{^5\negthinspace\log}2&=ab\\ ^3\negthinspace\log2&=ab \end{aligned}\)
\(\begin{aligned} 5^b&=2\\ ^5\negthinspace\log2&=b \end{aligned}\)
\(\begin{aligned} ^3\negthinspace\log5\cdot{^5\negthinspace\log}2&=ab\\ ^3\negthinspace\log2&=ab \end{aligned}\)
\(\begin{aligned}
{^{15}\negthinspace\log}40&=\dfrac{^3\negthinspace\log40}{^3\negthinspace\log15}\\[3.7pt]
&=\dfrac{^3\negthinspace\log\left(2^3\cdot5\right)}{^3\negthinspace\log(3\cdot5)}\\[3.7pt]
&=\dfrac{^3\negthinspace\log2^3+{^3\negthinspace\log}5}{^3\negthinspace\log3+{^3\negthinspace\log}5}\\[3.7pt]
&=\dfrac{3\ {^3\negthinspace\log}2+{^3\negthinspace\log}5}{^3\negthinspace\log3+{^3\negthinspace\log}5}\\
&=\boxed{\boxed{\dfrac{3ab+a}{1+a}}}
\end{aligned}\)
Jadi, \({^{15}\negthinspace\log}40=\dfrac{3ab+a}{1+a}\).
JAWAB: D
JAWAB: D
No.
Jika- 3,690
- −0,41
- 3,23
- −1,77
- 4,23
ALTERNATIF PENYELESAIAN
\begin{aligned}
^3\negmedspace\log 27x&={^3\negmedspace\log}27+{^3\negmedspace\log}x\\
&=3+1{,}23\\
&=\boxed{\boxed{4{,}23}}
\end{aligned}
Jadi, 3log 27x = 4,23.
JAWAB: E
JAWAB: E
No.
- −2
- −3
- 2
- 3
- −1
ALTERNATIF PENYELESAIAN
\begin{aligned}
^5\negmedspace\log 9+{^5\negmedspace\log 2}-{^5\negmedspace\log 450}&={^5\negmedspace\log\dfrac{9\cdot2}{450}}\\[3.7pt]
&={^5\negmedspace\log\dfrac1{25}}\\[3.7pt]
&={^5\negmedspace\log\dfrac1{5^2}}\\[3.7pt]
&={^5\negmedspace\log5^{-2}}\\
&=\boxed{\boxed{-2}}
\end{aligned}
Jadi, 5log 9 + 5log 2 − 5log 450 = −2.
JAWAB: A
JAWAB: A
No.
HitunglahALTERNATIF PENYELESAIAN
\begin{aligned}
{^2\negmedspace\log 48}-{^2\negmedspace\log 6}&={^2\negmedspace\log \dfrac{48}6}\\
&={^2\negmedspace\log 8}\\
&=\boxed{\boxed{3}}
\end{aligned}
Jadi, 2log 48 − 2log 6 = 3.
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