Exercise Zone : Logaritma

Berikut ini adalah kumpulan soal mengenai logaritma. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Jika ${^5\negmedspace\log 3 = x}$ dan ${^3\negmedspace\log 2 = y}$, maka $^{18}\negmedspace\log 15$ sama dengan

1. $\dfrac{2+x}{x(1+y)}$
2. $\dfrac{1+x}{x(2+y)}$
3. $\dfrac{2+y}{y(1+x)}$

4. $\dfrac{1+y}{y(2+x)}$
5. $\dfrac{3+x}{x(1+y)}$

$5log⁡3⋅3log⁡2=xy5log⁡2=xy$

$18log⁡15=5log⁡155log⁡18=5log⁡(5⋅3)5log⁡(32⋅2)=5log⁡5+5log⁡35log⁡32+5log⁡2=5log⁡5+5log⁡32 5log⁡3+5log⁡2=1+x2x+xy=1+xx(2+y)$

Jadi, $^{18}\negmedspace\log 15=\dfrac{1+x}{x(2+y)}$.
JAWAB: B

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No. 2

Jika $\dfrac{2-6\sqrt2}{\sqrt2-6}=x$, maka ${^x\negmedspace\log}\ 0{,}125=$ ....

$x=2−622−6⋅2+62+6=22+12−12−3622−36=−342−34=2$

$\eqalign{
{^x\negmedspace\log}\ 0{,}125&={^{\sqrt2}\negmedspace\log}\ \dfrac18\
&={^{2^{\frac12}}\negmedspace\log}\ 2^{-3}\
&=\dfrac21\cdot(-3)\
&=-6
}$

Jadi, ${^x\negmedspace\log}\ 0{,}125=-6$.

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No. 2

Jika $a\gt1$, $b\gt1$, dan $c\gt1$, maka $\left({^a\negthinspace\log}\dfrac1b\right)\left({^b\negthinspace\log}\dfrac1c\right)\left({^c\negthinspace\log}\dfrac1a\right)=$ ....

1. $1-abc$
2. $abc$

3. $-abc$
4. $1$

5. $-1$

$(alog1b)(blog1c)(clog1a)=(alogb−1)(blogc−1)(cloga−1)=(−1)(−1)(−1)(alogb)(blogc)(cloga)=−1$

Jadi, $\left({^a\negthinspace\log}\dfrac1b\right)\left({^b\negthinspace\log}\dfrac1c\right)\left({^c\negthinspace\log}\dfrac1a\right)=-1$.
JAWAB: E

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No. 3

Nilai dari $^{25}\negthinspace\log\dfrac1{64}\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8$ adalah ....

1. $-\dfrac32$
2. $-\dfrac12$
   
3. $\dfrac12$
   

4. $1$
5. $\dfrac32$
   

$\eqalign{
^{25}\negthinspace\log\dfrac1{64}\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8&=^{25}\negthinspace\log4^{-3}\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8\
&=-3\cdot{^{25}\negthinspace\log}4\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8\
&=-3\cdot{^{25}\negthinspace\log}10+{^{25}\negthinspace\log}8\
&={^{25}\negthinspace\log}10^{-3}+{^{25}\negthinspace\log}8\
&={^{25}\negthinspace\log}\dfrac1{1000}+{^{25}\negthinspace\log}8\
&={^{25}\negthinspace\log}\left(\dfrac1{1000}\cdot8\right)\
&={^{25}\negthinspace\log}\dfrac8{1000}\
&={^{25}\negthinspace\log}\dfrac1{125}\
&={^{5^2}\negthinspace\log}5^{-3}\
&=\boxed{\boxed{-\dfrac32}}
}$

Jadi, $^{25}\negthinspace\log\dfrac1{64}\cdot{^4\negthinspace\log}10+{^{25}\negthinspace\log}8=-\dfrac32$.
JAWAB: A

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No. 5

Sederhanakanlah

$\log 100+\log0,1$

Grup Whatsapp

$log⁡100+log⁡0,1=log⁡(100⋅0,1)=log⁡10=1$

Jadi, $\log 100+\log0,1=1$.

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No. 6

Sederhanakanlah $^3\negthinspace\log9-{^3\negthinspace\log}\dfrac13$

Grup Whatsapp

$3log⁡9−3log13=3log⁡913=3log⁡27=3$

Jadi, $^3\negthinspace\log9-{^3\negthinspace\log}\dfrac13=3$.

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No. 7

Sederhanakanlah
$^a\negthinspace\log1+{^a\negthinspace\log1}$

Grup Whatsapp

$alog⁡1+alog⁡1=0+0=0$

Jadi, $^a\negthinspace\log1+{^a\negthinspace\log1}=0$.

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No. 8

Sederhanakanlah
$^2\negthinspace\log\dfrac18+{^2\negthinspace\log\dfrac1{64}}$

Grup Whatsapp

$2log⁡18+2log⁡164=2log⁡123+2log⁡126=2log⁡2−3+2log⁡2−6=−3+(−6)=−9$

Jadi, $^2\negthinspace\log\dfrac18+{^2\negthinspace\log\dfrac1{64}}=-9$.

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No. 9

Sederhanakanlah
$^2\negthinspace\log2^8+{^2\negthinspace\log\dfrac18}$

Grup Whatsapp

$2log⁡28+2log⁡18=8+2log⁡12−3=8+(−3)=5$

Jadi, $^2\negthinspace\log2^8+{^2\negthinspace\log\dfrac18}=5$.

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No. 10

Sederhanakanlah
$^2\negthinspace\log16^{\frac12}+{^2\negthinspace\log8^{\frac13}}$

Grup Whatsapp

$2log⁡1612+2log⁡813=2log⁡(24)12+2log⁡(23)13=2log⁡22+2log⁡2=2+1=3$

Jadi, $^2\negthinspace\log16^{\frac12}+{^2\negthinspace\log8^{\frac13}}=3$.

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