Soal 11.
Jika ${^{27} \log {5}}=p$, ${^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=....$
Jawab: B.
${^{27} \log {5}}=p$
$\leftrightarrow {^{3^3} \log {5}}=p$
$\leftrightarrow \frac{1}{3}\cdot{^{3} \log {5}}=p$
$\leftrightarrow {^{3} \log {5}}=3p$ .........(pers. 1)
$\leftrightarrow {^{5} \log {3}}=\frac{1}{3p}$ .........(pers. 2)
${^{25} \log {3}}+{^{243} \log {\sqrt{5}}}={^{5^2} \log {3}}+{^{3^5} \log {5^{\frac{1}{2}}}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot{^{5} \log {3}}+\frac{\frac{1}{2}}{5} \cdot {^{3^5} \log {5^{\frac{1}{2}}}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot{^{5} \log {3}}+\frac{1}{10} \cdot {^{3} \log {5}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot{^{5} \log {3}}+\frac{1}{10} \cdot {^{3} \log {5}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot \frac{1}{3p}+\frac{1}{10} \cdot 3p$, (substitusikan pers. 1 dan pers. 2)
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{6p}+\frac{3p}{10}$
Soal 12.
Nilai dari $\frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}$
Jawab: D
$\frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=\frac{{^{3} \log {3^{-2}}}+{^{2^{1/2}} \log {3^{2}}}\cdot {^{3} \log {2^4}}}{^{2} \log {\frac{10}{5}}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=\frac{{-2}\cdot {^{3} \log {3}}+\frac{2}{\frac{1}{2}} \cdot {^{2} \log {3}}\cdot 4 \cdot {^{3} \log {2}}}{^{2} \log {2}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=\frac{{-2}\cdot 1+4 \cdot {^{2} \log {3}}\cdot 4 \cdot {^{3} \log {2}}}{1}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}={-2} + 4 \cdot 4 \cdot {^{2} \log {3}} \cdot {^{3} \log {2}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}={-2} + 16 \cdot {^{2} \log {2}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}={-2} + 4 \cdot 4 \cdot 1$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=14.$
Soal 13.
Nilai dari $\frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=....$
Jawab: D
$\frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{{^{3} \log {2}} \cdot {^{2^2} \log {3^3}} \cdot {^{3} \log {3^4}}}{{^{2} \log {\frac{8}{4}}}}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{{^{3} \log {2}} \cdot \frac{3}{2} \cdot {^{2} \log {3}} \cdot 4 \cdot {^{3} \log {3}}}{{^{2} \log {\frac{8}{4}}}}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{\frac{3}{2} \cdot 4 \cdot {^{3} \log {2}} \cdot {^{2} \log {3}} \cdot {^{3} \log {3}}}{{^{2} \log {2}}}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{6 \cdot {^{3} \log {3}} \cdot {1}}{1}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=6 \cdot 1$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=6$
Soal 14.
Nilai $x$ yang memenuhi ${^{\frac{1}{3}} \log {(x+ \sqrt {3})}}+{^{\frac{1}{3}} \log {(x- \sqrt {3})}} > 0=....$
Jawab: C
(1) ${^{\frac{1}{3}} \log {(x+ \sqrt {3})}}+{^{\frac{1}{3}} \log {(x- \sqrt {3})}} > 0$
$ \leftrightarrow {^{\frac{1}{3}} \log {(x+ \sqrt {3})(x- \sqrt {3})}} > {^{\frac{1}{3}} \log {1}}$
$ \leftrightarrow (x+ \sqrt {3})(x- \sqrt {3}) < 1$
$ \leftrightarrow x^2 - 3 < 1$
$ \leftrightarrow x^2 - 4 < 0 $
$ \leftrightarrow x^2 - 4 < 0$
$ \leftrightarrow (x+2)(x-2) < 0$
Pembuat nol:
$x=-2$ atau $x=2$
Garis Bilangan:
$-2<x<2$
Syarat:
(2) $x+ \sqrt {3}>0 \leftrightarrow x>-\sqrt{3}$
(3) $x- \sqrt {3}>0 \leftrightarrow x>\sqrt{3}$
Irisan dari ketiga solusi tersebut yaitu:
Jadi, himpunan penyelesaiannya yaitu $\sqrt{3}<x<2$ (memenuhi ketiga arsiran).
Soal 15.
Nilai dari $\frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=....$
Jawab: B
$\frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{{^{3^{\frac{1}{2}}} \log {10^2}} \cdot {\log {3^2}} - {^5 \log {5^4}}}{^2 \log {\frac{12}{3}}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{\frac{2}{\frac{1}{2}} \cdot {^{3} \log {10}} \cdot 2 \cdot {\log {3}} - 4 \cdot {^5 \log {5}}}{^2 \log {4}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{4 \cdot 2 \cdot {^{3} \log {10}} \cdot {\log {3}} - 4 \cdot 1}{^2 \log {2^2}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{8 \cdot {^{3} \log {3}} - 4}{2 \cdot {^2 \log {2}}}$, ingat bahwa ${\log {3}}$ memiliki basis 10 atau sama saja seperti ${^{10}\log {3}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{8 \cdot 1 - 4}{2 \cdot 1}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{4}{2}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=2$
Jika ${^{27} \log {5}}=p$, ${^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=....$
Jawab: B.
${^{27} \log {5}}=p$
$\leftrightarrow {^{3^3} \log {5}}=p$
$\leftrightarrow \frac{1}{3}\cdot{^{3} \log {5}}=p$
$\leftrightarrow {^{3} \log {5}}=3p$ .........(pers. 1)
$\leftrightarrow {^{5} \log {3}}=\frac{1}{3p}$ .........(pers. 2)
${^{25} \log {3}}+{^{243} \log {\sqrt{5}}}={^{5^2} \log {3}}+{^{3^5} \log {5^{\frac{1}{2}}}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot{^{5} \log {3}}+\frac{\frac{1}{2}}{5} \cdot {^{3^5} \log {5^{\frac{1}{2}}}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot{^{5} \log {3}}+\frac{1}{10} \cdot {^{3} \log {5}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot{^{5} \log {3}}+\frac{1}{10} \cdot {^{3} \log {5}}$
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{2}\cdot \frac{1}{3p}+\frac{1}{10} \cdot 3p$, (substitusikan pers. 1 dan pers. 2)
$\leftrightarrow {^{25} \log {3}}+{^{243} \log {\sqrt{5}}}=\frac{1}{6p}+\frac{3p}{10}$
Soal 12.
Nilai dari $\frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}$
Jawab: D
$\frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=\frac{{^{3} \log {3^{-2}}}+{^{2^{1/2}} \log {3^{2}}}\cdot {^{3} \log {2^4}}}{^{2} \log {\frac{10}{5}}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=\frac{{-2}\cdot {^{3} \log {3}}+\frac{2}{\frac{1}{2}} \cdot {^{2} \log {3}}\cdot 4 \cdot {^{3} \log {2}}}{^{2} \log {2}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=\frac{{-2}\cdot 1+4 \cdot {^{2} \log {3}}\cdot 4 \cdot {^{3} \log {2}}}{1}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}={-2} + 4 \cdot 4 \cdot {^{2} \log {3}} \cdot {^{3} \log {2}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}={-2} + 16 \cdot {^{2} \log {2}}$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}={-2} + 4 \cdot 4 \cdot 1$
$\leftrightarrow \frac{{^{3} \log {\frac{1}{9}}}+{^{\sqrt{2}} \log {9}}\cdot {^{3} \log {16}}}{{^{2} \log {10}}-{^{2} \log {5}}}=14.$
Soal 13.
Nilai dari $\frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=....$
Jawab: D
$\frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{{^{3} \log {2}} \cdot {^{2^2} \log {3^3}} \cdot {^{3} \log {3^4}}}{{^{2} \log {\frac{8}{4}}}}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{{^{3} \log {2}} \cdot \frac{3}{2} \cdot {^{2} \log {3}} \cdot 4 \cdot {^{3} \log {3}}}{{^{2} \log {\frac{8}{4}}}}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{\frac{3}{2} \cdot 4 \cdot {^{3} \log {2}} \cdot {^{2} \log {3}} \cdot {^{3} \log {3}}}{{^{2} \log {2}}}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=\frac{6 \cdot {^{3} \log {3}} \cdot {1}}{1}$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=6 \cdot 1$
$\leftrightarrow \frac{{^{3} \log {2}} \cdot {^{4} \log {27}} \cdot {^{3} \log {81}}}{{^{2} \log {8}} - {^{2} \log {4}}}=6$
Soal 14.
Nilai $x$ yang memenuhi ${^{\frac{1}{3}} \log {(x+ \sqrt {3})}}+{^{\frac{1}{3}} \log {(x- \sqrt {3})}} > 0=....$
Jawab: C
(1) ${^{\frac{1}{3}} \log {(x+ \sqrt {3})}}+{^{\frac{1}{3}} \log {(x- \sqrt {3})}} > 0$
$ \leftrightarrow {^{\frac{1}{3}} \log {(x+ \sqrt {3})(x- \sqrt {3})}} > {^{\frac{1}{3}} \log {1}}$
$ \leftrightarrow (x+ \sqrt {3})(x- \sqrt {3}) < 1$
$ \leftrightarrow x^2 - 3 < 1$
$ \leftrightarrow x^2 - 4 < 0 $
$ \leftrightarrow x^2 - 4 < 0$
$ \leftrightarrow (x+2)(x-2) < 0$
Pembuat nol:
$x=-2$ atau $x=2$
Garis Bilangan:
Syarat:
(2) $x+ \sqrt {3}>0 \leftrightarrow x>-\sqrt{3}$
(3) $x- \sqrt {3}>0 \leftrightarrow x>\sqrt{3}$
Irisan dari ketiga solusi tersebut yaitu:
Jadi, himpunan penyelesaiannya yaitu $\sqrt{3}<x<2$ (memenuhi ketiga arsiran).
Soal 15.
Nilai dari $\frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=....$
Jawab: B
$\frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{{^{3^{\frac{1}{2}}} \log {10^2}} \cdot {\log {3^2}} - {^5 \log {5^4}}}{^2 \log {\frac{12}{3}}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{\frac{2}{\frac{1}{2}} \cdot {^{3} \log {10}} \cdot 2 \cdot {\log {3}} - 4 \cdot {^5 \log {5}}}{^2 \log {4}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{4 \cdot 2 \cdot {^{3} \log {10}} \cdot {\log {3}} - 4 \cdot 1}{^2 \log {2^2}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{8 \cdot {^{3} \log {3}} - 4}{2 \cdot {^2 \log {2}}}$, ingat bahwa ${\log {3}}$ memiliki basis 10 atau sama saja seperti ${^{10}\log {3}}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{8 \cdot 1 - 4}{2 \cdot 1}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=\frac{4}{2}$
$\leftrightarrow \frac{{^{\sqrt {3}} \log {100}} \cdot {\log {9}} - {^5 \log {625}}}{{^2 \log {12}} - {^2 \log {3}}}=2$
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