HOTS Zone : Logaritma
Table of Contents
Tipe:
No. 1
\(\dfrac{^4\negmedspace\log20^2}{{^4\negmedspace\log^280}-{^4\negmedspace\log^25}}=\) ....- −1
- \(-\dfrac12\)
- \(\dfrac12\)
- 1
- 2
ALTERNATIF PENYELESAIAN
\(\eqalign{
\dfrac{^4\negmedspace\log20^2}{{^4\negmedspace\log^280}-{^4\negmedspace\log^25}}&=\dfrac{^4\negmedspace\log20^2}{\left({^4\negmedspace\log80}+{^4\negmedspace\log5}\right)\left({^4\negmedspace\log80}-{^4\negmedspace\log5}\right)}\\
&=\dfrac{^4\negmedspace\log20^2}{{^4\negmedspace\log(80\cdot5)}\cdot{^4\negmedspace\log\dfrac{80}5}}\\
&=\dfrac{^4\negmedspace\log20^2}{{^4\negmedspace\log400}\cdot{^4\negmedspace\log16}}\\
&=\dfrac{^4\negmedspace\log20^2}{{^4\negmedspace\log20^2}\cdot2}\\
&=\boxed{\boxed{\dfrac12}}
}\)
Jadi, \(\dfrac{^4\negmedspace\log20^2}{{^4\negmedspace\log^280}-{^4\negmedspace\log^25}}=\dfrac12\).
JAWAB: C
JAWAB: C
No. 2
Bentuk \((c)^{^{ab}\negmedspace\log b}\cdot(abc)^{^{ab}\negmedspace\log a}\) dapat disederhanakan menjadi ....- a
- b
- c
- ac
- abc
ALTERNATIF PENYELESAIAN
\(\eqalign{
(c)^{^{ab}\negthinspace\log b}\cdot(abc)^{^{ab}\negthinspace\log a}&=(c)^{^{ab}\negthinspace\log b}\cdot(ab)^{^{ab}\negthinspace\log a}\cdot(c)^{^{ab}\negthinspace\log a}\\
&=(c)^{^{ab}\negthinspace\log b+{^{ab}\negthinspace\log a}}\cdot a\\
&=(c)^{^{ab}\negthinspace\log ab}\cdot a\\
&=c\cdot a\\
&=\boxed{\boxed{ac}}
}\)
Jadi, \((c)^{^{ab}\negmedspace\log b}\cdot(abc)^{^{ab}\negmedspace\log a}=ac\).
JAWAB: D
JAWAB: D
No.
Jika diketahui nilai- 2
- 3
- 4
- 5
- 6
ALTERNATIF PENYELESAIAN
Misal 2log a = x, 2log b = y, 2log c = z
\(\begin{aligned} a\cdot b\cdot c&=2^6\\ ^2\negmedspace\log(a\cdot b\cdot c)&=6\\ {^2\negmedspace\log a}+{^2\negmedspace\log b}+{^2\negmedspace\log c}&=6\\ x+y+z&=6 \end{aligned}\)
\(\begin{aligned} \left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log bc}\right)+\left({^2\negmedspace\log b}\right)\left({^2\negmedspace\log c}\right)&=10\\ \left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log b}+{^2\negmedspace\log c}\right)+\left({^2\negmedspace\log b}\right)\left({^2\negmedspace\log c}\right)&=10\\ \left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log b}\right)+\left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log c}\right)+\left({^2\negmedspace\log b}\right)\left({^2\negmedspace\log c}\right)&=10\\ xy+xz+yz&=10 \end{aligned}\)
\(\begin{aligned} \sqrt{{^2\negmedspace\log^2 a}+{^2\negmedspace\log^2b}+{^2\negmedspace\log^2c}}&=\sqrt{x^2+y^2+z^2}\\ &=\sqrt{(x+y+z)^2-2(xy+xz+yz)}\\ &=\sqrt{6^2-2(10)}\\ &=\sqrt{16}\\ &=\boxed{\boxed{4}} \end{aligned}\)
\(\begin{aligned} a\cdot b\cdot c&=2^6\\ ^2\negmedspace\log(a\cdot b\cdot c)&=6\\ {^2\negmedspace\log a}+{^2\negmedspace\log b}+{^2\negmedspace\log c}&=6\\ x+y+z&=6 \end{aligned}\)
\(\begin{aligned} \left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log bc}\right)+\left({^2\negmedspace\log b}\right)\left({^2\negmedspace\log c}\right)&=10\\ \left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log b}+{^2\negmedspace\log c}\right)+\left({^2\negmedspace\log b}\right)\left({^2\negmedspace\log c}\right)&=10\\ \left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log b}\right)+\left({^2\negmedspace\log a}\right)\left({^2\negmedspace\log c}\right)+\left({^2\negmedspace\log b}\right)\left({^2\negmedspace\log c}\right)&=10\\ xy+xz+yz&=10 \end{aligned}\)
\(\begin{aligned} \sqrt{{^2\negmedspace\log^2 a}+{^2\negmedspace\log^2b}+{^2\negmedspace\log^2c}}&=\sqrt{x^2+y^2+z^2}\\ &=\sqrt{(x+y+z)^2-2(xy+xz+yz)}\\ &=\sqrt{6^2-2(10)}\\ &=\sqrt{16}\\ &=\boxed{\boxed{4}} \end{aligned}\)
Jadi, \(\sqrt{{^2\negmedspace\log^2 a}+{^2\negmedspace\log^2b}+{^2\negmedspace\log^2c}}=4\).
JAWAB: C
JAWAB: C
No.
Diberikan fungsi \(f(n)=1+\displaystyle\sum_{k=1}^n\left\lfloor{^3\negthinspace\log k}\right\rfloor\) yang terdefinisi pada bilangan asli n. Jika p adalah suatu bilangan sehingga- 440 ≤ p < 450
- 450 ≤ p < 460
- 460 ≤ p < 470
- 470 ≤ p < 480
- 480 ≤ p < 490
ALTERNATIF PENYELESAIAN
Misal a adalah bilangan bulat tak-negatif. Untuk 3a ≤ k ≤ 3a + 1 − 1,
⌊3log k⌋ = a
\(\begin{aligned} \displaystyle\sum_{k\ =\ 3^a}^{3^{a+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=a\left(3^{a+1}-1-3^a+1\right)\\ &=a\left(3\cdot3^a-3^a\right)\\ &=2a3^a \end{aligned}\)
\(\begin{aligned} \displaystyle\sum_{k\ =\ 1\ =\ 3^0}^{n\ =\ 3^{b+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2\displaystyle\sum_{a=0}^{b}a3^a\\ &=2\left(\dfrac{3-(b+1)3^{b+1}+b3^{b+2}}{(1-3)^2}\right)\\[4pt] &=2\left(\dfrac{3-(b+1)3^{b+1}+3b3^{b+1}}{4}\right)\\[4pt] &=\dfrac{3+(2b-1)3^{b+1}}2 \end{aligned}\)
Kita car nilai b sehingga \(\dfrac{3+(2b-1)3^{b+1}}2\) mendekati atau sama dengan 2022
\(\begin{aligned} \dfrac{3+(2b-1)3^{b+1}}2&\leq2022\\[4pt] 3+(2b-1)3^{b+1}&\leq4044\\ (2b-1)3^{b+1}&\leq4041\\ (2b-1)3^b&\leq1347 \end{aligned}\)
Jika b = 5,
(2(5) − 1)35 = 2187 > 1347
Jika b = 4,
(2(4) − 1)34 = 567 < 1347
didapat b = 4
\(\dfrac{3+(2(4)-1)3^{4+1}}2=852\)
\(\begin{aligned} f(p)&=2023\\ 1+\displaystyle\sum_{k=1}^p\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2023\\ \displaystyle\sum_{k=1}^{3^{4+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor+\displaystyle\sum_{k\ =\ 3^5\ =\ 243}^{p}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2022\\ 852+5(p-243+1)&=2022\\ p&=476 \end{aligned}\)
⌊3log k⌋ = a
\(\begin{aligned} \displaystyle\sum_{k\ =\ 3^a}^{3^{a+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=a\left(3^{a+1}-1-3^a+1\right)\\ &=a\left(3\cdot3^a-3^a\right)\\ &=2a3^a \end{aligned}\)
\(\begin{aligned} \displaystyle\sum_{k\ =\ 1\ =\ 3^0}^{n\ =\ 3^{b+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2\displaystyle\sum_{a=0}^{b}a3^a\\ &=2\left(\dfrac{3-(b+1)3^{b+1}+b3^{b+2}}{(1-3)^2}\right)\\[4pt] &=2\left(\dfrac{3-(b+1)3^{b+1}+3b3^{b+1}}{4}\right)\\[4pt] &=\dfrac{3+(2b-1)3^{b+1}}2 \end{aligned}\)
Kita car nilai b sehingga \(\dfrac{3+(2b-1)3^{b+1}}2\) mendekati atau sama dengan 2022
\(\begin{aligned} \dfrac{3+(2b-1)3^{b+1}}2&\leq2022\\[4pt] 3+(2b-1)3^{b+1}&\leq4044\\ (2b-1)3^{b+1}&\leq4041\\ (2b-1)3^b&\leq1347 \end{aligned}\)
Jika b = 5,
(2(5) − 1)35 = 2187 > 1347
Jika b = 4,
(2(4) − 1)34 = 567 < 1347
didapat b = 4
\(\dfrac{3+(2(4)-1)3^{4+1}}2=852\)
\(\begin{aligned} f(p)&=2023\\ 1+\displaystyle\sum_{k=1}^p\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2023\\ \displaystyle\sum_{k=1}^{3^{4+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor+\displaystyle\sum_{k\ =\ 3^5\ =\ 243}^{p}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2022\\ 852+5(p-243+1)&=2022\\ p&=476 \end{aligned}\)
Jadi, nilai p yang memenuhi berada pada interval 470 ≤ p < 480.
JAWAB: D
JAWAB: D
No.
Diberikan bilangan bulat positif a, b, c, dan d yang memenuhi \begin{aligned} \log_ab&=\dfrac32\\[4pt] \log_cd&=\dfrac54\\[4pt] a-c&=9 \end{aligned} Misalkan- 10
- 12
- 11
- 9
- 16
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\log_ab&=\dfrac32\\[4pt]
a^3&=b^2=m^6
\end{aligned}\)
a = m2 , b = m3
\(\begin{aligned} \log_cd&=\dfrac54\\[4pt] c^5&=d^4=n^{20} \end{aligned}\)
c = n4 , d = n5
\(\begin{aligned} a-c&=9\\ m^2-n^4&=9\\ \left(m+n^2\right)\left(m-n^2\right)&=9\cdot1=3\cdot3 \end{aligned}\)
Karenam + n2 > m − n2, maka
\(\begin{aligned} m+n^2&=9\\ m-n^2&=1&\ +\\\hline 2m&=10\\ m&=5 \end{aligned}\)
Didapat,n = 2
b = 53 = 125
d = 25 = 32
125 − 32 = 93
9 + 3 =12
\(\begin{aligned} \log_cd&=\dfrac54\\[4pt] c^5&=d^4=n^{20} \end{aligned}\)
\(\begin{aligned} a-c&=9\\ m^2-n^4&=9\\ \left(m+n^2\right)\left(m-n^2\right)&=9\cdot1=3\cdot3 \end{aligned}\)
Karena
\(\begin{aligned} m+n^2&=9\\ m-n^2&=1&\ +\\\hline 2m&=10\\ m&=5 \end{aligned}\)
Didapat,
125 − 32 = 93
9 + 3 =
Jadi, jumlahan dari digit-digit dari X adalah 12.
JAWAB: B
JAWAB: B
No.
Jika $\dfrac{\log x}{q-r}=\dfrac{\log y}{r-p}=\dfrac{\log z}{p-q}$, maka nilai dari $x^{q+r}\cdot y^{r+p}\cdot z^{p+q}$ adalah ....- 1
- −1
- 0
- 2
- −2
ALTERNATIF PENYELESAIAN
Misal $\dfrac{\log x}{q-r}=\dfrac{\log y}{r-p}=\dfrac{\log z}{p-q}=k$
$\log x=k(q-r)$, $\log y=k(r-p)$, $\log z=k(p-q)$
$x=10^{k(q-r)}$, $y=10^{k(r-p)}$, $z=10^{k(p-q)}$
\(\begin{aligned} x^{q+r}\cdot y^{r+p}\cdot z^{p+q}&=10^{k(q-r)(q+r)}\cdot10^{k(r-p)(r+p)}\cdot10^{k(p-q)(p+q)}\\ &=10^{k(q^2-r^2)}\cdot10^{k(r^2-p^2)}\cdot10^{k(p^2-q^2)}\\ &=10^{k(q^2-r^2+r^2-p^2+p^2-q^2)}\\ &=10^{k(0)}\\ &=\boxed{\boxed{1}} \end{aligned}\)
$\log x=k(q-r)$, $\log y=k(r-p)$, $\log z=k(p-q)$
$x=10^{k(q-r)}$, $y=10^{k(r-p)}$, $z=10^{k(p-q)}$
\(\begin{aligned} x^{q+r}\cdot y^{r+p}\cdot z^{p+q}&=10^{k(q-r)(q+r)}\cdot10^{k(r-p)(r+p)}\cdot10^{k(p-q)(p+q)}\\ &=10^{k(q^2-r^2)}\cdot10^{k(r^2-p^2)}\cdot10^{k(p^2-q^2)}\\ &=10^{k(q^2-r^2+r^2-p^2+p^2-q^2)}\\ &=10^{k(0)}\\ &=\boxed{\boxed{1}} \end{aligned}\)
Jadi, $x^{q+r}\cdot y^{r+p}\cdot z^{p+q}=1$.
JAWAB: A
JAWAB: A
No.
Diketahui- 1
- 2
- 3
- 4
- 5
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
^2\negmedspace\log(x-1)+{^2\negmedspace\log(x+2)}&=3\\
^2\negmedspace\log\left((x-1)(x+2)\right)&=3\\
^2\negmedspace\log(x^2+x-2)&=3\\
x^2+x-2&=2^3\\
x^2+x-2&=8\\
x^2+x&=10
\end{aligned}\)
\(\begin{aligned} ^2\negmedspace\log(x-2)+{^2\negmedspace\log(x+3)}&=^2\negmedspace\log\left((x-2)(x+3)\right)\\ &=^2\negmedspace\log(x^2+x-6)\\ &=^2\negmedspace\log(10-6)\\ &=^2\negmedspace\log4\\ &=2 \end{aligned}\)
\(\begin{aligned} ^2\negmedspace\log(x-2)+{^2\negmedspace\log(x+3)}&=^2\negmedspace\log\left((x-2)(x+3)\right)\\ &=^2\negmedspace\log(x^2+x-6)\\ &=^2\negmedspace\log(10-6)\\ &=^2\negmedspace\log4\\ &=2 \end{aligned}\)
Jadi, nilai dari 2log(x − 2) + 2log(x + 3) = 2 .
JAWAB: B
JAWAB: B
No.
log2(x − 1) + log2(x + 2) = 3. Nilai dariALTERNATIF PENYELESAIAN
\(\begin{aligned}
\log_2(x-1)+\log_2(x+2)&=3\\
\log_2\left(x^2+x-2\right)&=3\\
x^2+x-2&=8\\
x^2+x&=10
\end{aligned}\)
\(\begin{aligned} \log_2(x-2)+\log_2(x+3)&=\log_2\left(x^2+x-6\right)\\ &=\log_2\left(10-6\right)\\ &=\log_24\\ &=\color{blue}\boxed{\boxed{\color{black}2}} \end{aligned}\)
\(\begin{aligned} \log_2(x-2)+\log_2(x+3)&=\log_2\left(x^2+x-6\right)\\ &=\log_2\left(10-6\right)\\ &=\log_24\\ &=\color{blue}\boxed{\boxed{\color{black}2}} \end{aligned}\)
Jadi, log2(x − 2) + log2(x + 3) = 2 .
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