Berikut ini adalah kumpulan soal mengenai logaritma. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
Jika
{^5\negmedspace\log 3 = x} dan
{^3\negmedspace\log 2 = y}, maka
^{18}\negmedspace\log 15 sama dengan
- \dfrac{2+x}{x(1+y)}
- \dfrac{1+x}{x(2+y)}
- \dfrac{2+y}{y(1+x)}
- \dfrac{1+y}{y(2+x)}
- \dfrac{3+x}{x(1+y)}
\begin{aligned}
^5\negmedspace\log 3\cdot{^3\negmedspace\log 2}&=xy\\
^5\negmedspace\log2&=xy
\end{aligned}
\begin{aligned}
^{18}\negmedspace\log 15&=\dfrac{^5\negmedspace\log15}{^5\negmedspace\log18}\\[8pt]
&=\dfrac{^5\negmedspace\log(5\cdot3)}{^5\negmedspace\log\left(3^2\cdot2\right)}\\[8pt]
&=\dfrac{^5\negmedspace\log5+{^5\negmedspace\log3}}{^5\negmedspace\log3^2+{^5\negmedspace\log2}}\\[8pt]
&=\dfrac{^5\negmedspace\log5+{^5\negmedspace\log3}}{2\ {^5\negmedspace\log3}+{^5\negmedspace\log2}}\\[8pt]
&=\dfrac{1+x}{2x+xy}\\
&=\boxed{\boxed{\dfrac{1+x}{x(2+y)}}}
\end{aligned}
No. 2
Jika \dfrac{2-6\sqrt2}{\sqrt2-6}=x, maka {^x\negmedspace\log}\ 0{,}125= ....\begin{aligned}
x&=\dfrac{2-6\sqrt2}{\sqrt2-6}\cdot\dfrac{\sqrt2+6}{\sqrt2+6}\\[4pt]
&=\dfrac{2\sqrt2+12-12-36\sqrt2}{2-36}\\[4pt]
&=\dfrac{-34\sqrt2}{-34}\\[4pt]
&=\sqrt2
\end{aligned}
$\eqalign{
{^x\negmedspace\log}\ 0{,}125&={^{\sqrt2}\negmedspace\log}\ \dfrac18\\
&={^{2^{\frac12}}\negmedspace\log}\ 2^{-3}\\
&=\dfrac21\cdot(-3)\\
&=-6
}$
No. 2
Jika a\gt1, b\gt1, dan c\gt1, maka \left({^a\negthinspace\log}\dfrac1b\right)\left({^b\negthinspace\log}\dfrac1c\right)\left({^c\negthinspace\log}\dfrac1a\right)= ....\begin{aligned}
\left({^a\negthinspace\log}\dfrac1b\right)\left({^b\negthinspace\log}\dfrac1c\right)\left({^c\negthinspace\log}\dfrac1a\right)&=\left({^a\negthinspace\log}b^{-1}\right)\left({^b\negthinspace\log}c^{-1}\right)\left({^c\negthinspace\log}a^{-1}\right)\\
&=(-1)(-1)(-1)\left({^a\negthinspace\log}b\right)\left({^b\negthinspace\log}c\right)\left({^c\negthinspace\log}a\right)\\
&=\boxed{\boxed{-1}}
\end{aligned}
No. 3
Nilai dari ^{25}\negthinspace\log\dfrac1{64}\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8 adalah ....- -\dfrac32
- -\dfrac12
- \dfrac12
$\eqalign{
^{25}\negthinspace\log\dfrac1{64}\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8&=^{25}\negthinspace\log4^{-3}\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8\\
&=-3\cdot{^{25}\negthinspace\log}4\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8\\
&=-3\cdot{^{25}\negthinspace\log}10+{^{25}\negthinspace\log}8\\
&={^{25}\negthinspace\log}10^{-3}+{^{25}\negthinspace\log}8\\
&={^{25}\negthinspace\log}\dfrac1{1000}+{^{25}\negthinspace\log}8\\
&={^{25}\negthinspace\log}\left(\dfrac1{1000}\cdot8\right)\\
&={^{25}\negthinspace\log}\dfrac8{1000}\\
&={^{25}\negthinspace\log}\dfrac1{125}\\
&={^{5^2}\negthinspace\log}5^{-3}\\
&=\boxed{\boxed{-\dfrac32}}
}$
No. 5
Sederhanakanlah\log 100+\log0,1
\begin{aligned}
\log 100+\log0,1&=\log (100\cdot0,1)\\
&=\log10\\
&=1
\end{aligned}
No. 6
Sederhanakanlah ^3\negthinspace\log9-{^3\negthinspace\log}\dfrac13\begin{aligned}
^3\negthinspace\log9-{^3\negthinspace\log}\dfrac13&={^3\negthinspace\log\dfrac9{\dfrac13}}\\[6pt]
&={^3\negthinspace\log27}\\
&=3
\end{aligned}
No. 7
Sederhanakanlah
^a\negthinspace\log1+{^a\negthinspace\log1}\begin{aligned}
^a\negthinspace\log1+{^a\negthinspace\log1}&=0+0\\
&=0
\end{aligned}
No. 8
Sederhanakanlah
^2\negthinspace\log\dfrac18+{^2\negthinspace\log\dfrac1{64}}\begin{aligned}
^2\negthinspace\log\dfrac18+{^2\negthinspace\log\dfrac1{64}}&={^2\negthinspace\log\dfrac1{2^3}}+{^2\negthinspace\log\dfrac1{2^6}}\\[4pt]
&={^2\negthinspace\log2^{-3}}+{^2\negthinspace\log2^{-6}}\\
&=-3+(-6)\\
&=-9
\end{aligned}
No. 9
Sederhanakanlah
^2\negthinspace\log2^8+{^2\negthinspace\log\dfrac18}\begin{aligned}
^2\negthinspace\log2^8+{^2\negthinspace\log\dfrac18}&=8+{^2\negthinspace\log\dfrac1{2^{-3}}}\\[8pt]
&=8+(-3)\\
&=5
\end{aligned}
No. 10
Sederhanakanlah
^2\negthinspace\log16^{\frac12}+{^2\negthinspace\log8^{\frac13}}\begin{aligned}
^2\negthinspace\log16^{\frac12}+{^2\negthinspace\log8^{\frac13}}&=^2\negthinspace\log\left(2^4\right)^{\frac12}+{^2\negthinspace\log\left(2^3\right)^{\frac13}}\\[5pt]
&=^2\negthinspace\log2^2+{^2\negthinspace\log2}\\
&=2+1\\
&=3
\end{aligned}
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