Lanjutan Contoh Soal (Bentuk Pangkat, Akar, dan Logaritma (KTSP)) 4

\fbox{31}.\quad \textrm{Diketahui}\,\: ^{7}\log 2=a\,\: \textrm{dan}\,\: ^{2}\log 3=b,\,\: \textrm{maka nilai} \,\: ^{98}\log 6\,\: \textrm{= ....}.

Jawab:

\begin{aligned}\textrm{Dari soal}&\: \, \textrm{diketahui}\\ &\\ &\begin{cases} & ^{7}\log 2=a \: \: \Leftrightarrow \: \: ^{2}\log 7=\displaystyle \frac{1}{a}\\ &\\ & ^{2}\log 3=b \: \: \Leftrightarrow \: \: ^{3}\log 2=\displaystyle \frac{1}{b} \end{cases}\\ &\\ ^{98}\log 6&=\displaystyle \frac{\log 6}{\log 98}\\ &=\: \displaystyle \frac{^{2}\log 2.3}{^{2}\log 2.7.7}\\ &=\: \displaystyle \frac{^{2}\log 2+\: ^{2}\log 3}{^{2}\log 2+\: ^{2}\log 7+\: ^{2}\log 7}\\ &=\displaystyle \frac{1+b}{1+\displaystyle \frac{1}{a}+\displaystyle \frac{1}{a}}\\ &=\displaystyle \frac{1+b}{\displaystyle \frac{a+2}{a}}\\ &=\displaystyle \frac{a(1+b)}{2+a} \end{aligned}.

\fbox{32}.\quad \textbf{(UN MAT IPA 2012)}\textrm{Diketahui}\,\: ^{5}\log 3=a\,\: \textrm{dan}\,\: ^{3}\log 4=b,\,\: \textrm{maka nilai} \,\: ^{4}\log 15\,\: \textrm{= ....}.

Jawab:

\begin{aligned}\textrm{Dari soal}&\: \, \textrm{diketahui}\\ &\begin{cases} & ^{5}\log 3=a \: \: \Leftrightarrow \: \: ^{3}\log 5=\displaystyle \frac{1}{a}\\ &\\ & ^{3}\log 4=b \: \: \Leftrightarrow \: \: ^{4}\log 3=\displaystyle \frac{1}{b} \end{cases}\\ &\\ ^{4}\log 15&=\displaystyle \frac{\log 15}{\log 4}\\ &=\: \displaystyle \frac{^{3}\log 3.5}{^{3}\log 4}\\ &=\: \displaystyle \frac{^{3}\log 3+\: ^{3}\log 5}{^{3}\log 4}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{a}}{b}\\ &=\displaystyle \frac{(a+1)}{ab} \end{aligned}.

\fbox{33}.\quad \textbf{(UN MAT IPA 2015)}\textrm{Hasil}\,\: \displaystyle \frac{^{6}\log 27.^{\sqrt{3}}\log 36-^{4}\log \displaystyle \frac{1}{64}}{^{6}\log 36-^{6}\log 6\sqrt{6}}\,\: \textrm{adalah = ....}.

Jawab:

\begin{aligned}\displaystyle \frac{^{6}\log 27.^{\sqrt{3}}\log 36-^{4}\log \displaystyle \frac{1}{64}}{^{6}\log 36-^{6}\log 6\sqrt{6}}&=\displaystyle \frac{^{6}\log 3^{3}.^{3^{\frac{1}{2}}}\log 6^{2}-^{4}\log 4^{-3}}{^{6}\log 6^{2}-^{6}\log 6^{\frac{3}{2}}}\\ &=\displaystyle \frac{3.\left ( \frac{2}{\frac{1}{2}}.^{6}\log 3.^{3}\log 6 \right )-(-3).^{4}\log 4}{2.^{6}\log 6-\frac{3}{2}.^{6}\log 6}\\ &=\displaystyle \frac{12+3}{2-\frac{3}{2}}\\ &=\displaystyle \frac{15}{\frac{1}{2}}\\ &=30 \end{aligned}.

\fbox{34}.\quad \textbf{(UN MAT IPA 2015)}\textrm{Hasil}\,\: \displaystyle \frac{^{7}\log 16\sqrt{2}.^{2\sqrt{2}}\log \displaystyle \frac{1}{49}+^{2}\log \displaystyle \frac{1}{16}}{^{5}\log 5\sqrt{5}+^{5}\log 25\sqrt{5}}\,\: \textrm{adalah = ....}.

Jawab:

\begin{aligned}\displaystyle \frac{^{7}\log 16\sqrt{2}.\: ^{2\sqrt{2}}\log \displaystyle \frac{1}{49}+\: ^{2}\log \displaystyle \frac{1}{16}}{^{5}\log 5\sqrt{5}+\: ^{5}\log 25\sqrt{5}}&=\displaystyle \frac{^{7}\log 2^{\frac{9}{2}}.\: ^{2^{\frac{3}{2}}}\log 7^{-2}+\: ^{2}\log 2^{-4}}{^{5}\log 5^{\frac{3}{2}}+\: ^{5}\log 5^{\frac{5}{2}}}\\ &=\displaystyle \frac{\frac{9}{2}.\left ( \frac{-2}{\frac{3}{2}} \right ).^{7}\log 2.^{2}\log 7+(-4).^{2}\log 2}{\frac{3}{2}.^{5}\log 5+\frac{5}{2}.^{5}\log 5}\\ &=\displaystyle \frac{-6-4}{\frac{3}{2}+\frac{5}{2}}\\ &=\displaystyle \frac{-10}{4}\\ &=-\displaystyle \frac{5}{2}\end{aligned}.

\fbox{35}.\quad \textbf{(UN MAT IPA 2015)}\textrm{Hasil}\,\: \displaystyle \frac{^{4}\log 5\sqrt{5}.^{25}\log 16+^{6}\log \displaystyle \frac{1}{216}}{^{4}\log 16\sqrt{2}+^{4}\log \displaystyle \frac{1}{32}}\,\: \textrm{adalah = ....}.

Jawab:

\begin{aligned}\displaystyle \frac{^{4}\log 5\sqrt{5}.\: ^{25}\log 16+\: ^{6}\log \displaystyle \frac{1}{216}}{^{4}\log 16\sqrt{2}+\: ^{4}\log \displaystyle \frac{1}{32}}&=\displaystyle \frac{^{4}\log 5^{\frac{3}{2}}.\: ^{5^{2}}\log 4^{2}+\: ^{6}\log 6^{-3}}{^{2^{2}}\log 2^{\frac{9}{2}}+\: ^{2^{2}}\log 2^{-5}}\\ &=\displaystyle \frac{\frac{3}{2}.\left ( \frac{2}{2} \right ).^{4}\log 5.^{5}\log 4+(-3).^{6}\log 6}{\frac{\frac{9}{2}}{2}.^{2}\log 2+\frac{-5}{2}.^{2}\log 2}\\ &=\displaystyle \frac{-\frac{3}{2}-3}{\frac{9}{4}-\frac{5}{2}}\\ &=\displaystyle \frac{-\frac{3}{2}}{-\frac{1}{4}}\\ &=\displaystyle 6\end{aligned}.

\begin{array}{ll}\\ \fbox{36}.&\textbf{(SPMU 2010 kode soal 4928)}\textrm{Jika}\,\: 2^{a+1}=6\,\: \textrm{dan}\, \: b=\: ^{3}\log 5\: \: \textrm{maka nilai} \: \: ^{2}\log 15\: \: \textrm{adalah = ....} \end{array}.

Jawab:

\begin{aligned}\textrm{Diketahui bahwa}&\\ &\begin{cases} \bullet \quad 2^{a+1}=6 & \Rightarrow 2^{a}.2=6\: \: \Rightarrow \: \: 2^{a}=3 \\ &\Rightarrow a =\: ^{2}\log 3 \: \: ,\textrm{dan}\: \: \displaystyle \frac{1}{a}=\: ^{3}\log 2\\ \bullet \quad b=\: ^{3}\log 5 & \end{cases}\\ &\\ ^{2}\log 15&=\displaystyle \frac{^{3}\log 15}{^{3}\log 2},\quad \textrm{angka 3 adalah yang paling mungkin dengan melihat yang diketahui}\\ &=\displaystyle \frac{^{3}\log 3.5}{\frac{1}{a}}\: \quad \textrm{sebagai bilangan dasar}\\ &=a\left ( ^{3}\log 3+\: ^{3}\log 5 \right )\\ &=a\left ( 1+b \right ) \end{aligned}.

\begin{array}{ll}\\ \fbox{37}.&\textbf{(SPMU 2010 kode soal 4928)}\textrm{Jika}\,\: a\,\: \textrm{memenuhi persamaan}\\ &\\ &\: ^{4}\log \displaystyle \frac{^{4}\log a}{^{4}\log \: ^{4}\log 16}=2\: ,\: \textrm{maka nilai dari} \: \: ^{2}\log a\: \: \textrm{adalah = ....} \end{array}.

Jawab:

\begin{aligned}^{4}\log \displaystyle \frac{^{4}\log a}{^{4}\log \: ^{4}\log 16}&=2\\ \displaystyle \frac{^{4}\log a}{^{4}\log \: (2)}&=4^{2}=16\\ ^{2}\log a&=16\\ a&=2^{16}\\ \textrm{Sehingga nilai}&\: \: \\ ^{2}\log a=\: ^{2}\log 2^{16}&=16 \end{aligned}.

\begin{array}{ll}\\ \fbox{38}.&\textbf{(SPMB 2006)}\textrm{Jika}\,\: ^{81}\log \displaystyle \frac{1}{x}=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log 81\,\:, \textrm{maka}\\ &\\ &\: \textrm{maka nilai dari} \: \: 2x-3y\: \: \textrm{adalah = ....} \end{array}.

Jawab:

\textrm{Perhatikan saja}\: \: ^{81}\log \displaystyle \frac{1}{x}=\: ^{x}\log \displaystyle \frac{1}{y}\\ \textrm{Jelas ini hanya dapat dipenuhi untuk} \: \: x=81 \\ \textrm{Karena} \: \: \displaystyle \frac{1}{x}=\displaystyle \frac{1}{y}\: \Rightarrow \: y=x=81\: \: \textrm{juga}.\\ \textrm{Sehingga}\: \: 2x-3y=2x-3x=-x=-81..

\begin{array}{ll}\\ \fbox{39}.&\textrm{Penyelesaian dari persamaan berikut}\\ &\\ &\: ^{2x-3}\log ^{-1}2.\: ^{x}\log 2-\: ^{x+6}\log ^{-1}x+\displaystyle \frac{1}{^{x+2}\log x}=1\: \: \textrm{adalah ....} \end{array}.

Jawab:

\begin{aligned}&^{2x-3}\log ^{-1}2.\: ^{x}\log 2-\: ^{x+6}\log ^{-1}x+\displaystyle \frac{1}{^{x+2}\log x}=1 \\ &^{x}\log 2.\: ^{2}\log (2x-3)-\: ^{x}\log (x+6)+\: ^{x}\log (x+2)=1,&\textnormal{ingat bahwa:}\: \: ^{a}\log^{-1} b=\: ^{b}\log a\\ &^{x}\log (2x-3)+\: ^{x}\log (x+2)-\: ^{x}\log (x+6)=1,&\textnormal{ingat juga}\: \: ^{a}\log b.\: ^{b}\log c=\: ^{a}\log c\\ &^{x}\log \displaystyle \frac{(2x-3)(x+2)}{(x+6)}=1\\ &\displaystyle \frac{(2x-3)(x+2)}{(x+6)}=x^{1}\\ &\displaystyle \frac{(2x^{2}+x-6)}{(x+6)}=x\\ &2x^{2}+x-6=x^{2}+6x\\ &x^{2}-5x-6=0\\ &(x-6)(x+1)=0\\ &x-6=0\: \: \textrm{atau}\: \: x+1=0\\ &x=6\: \textrm{(memenuhi)}\: \: \textrm{atau}\: \: x=-1\: \textrm{(tidak memenuhi)} \end{aligned}.

\begin{array}{ll}\\ \fbox{40}.&\textrm{Tentukanlah nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\: \: ^{2015}\log x^{x^{2015}}=1\end{array}.\\\\\\ \textrm{Jawab:}\\\\\\ \begin{aligned}^{2015}\log x^{x^{2015}}&=1\\ x^{x^{2015}}&=2015^{1}\\ x^{x^{2015}}&=2015\\ \left ( x^{x^{2015}} \right )^{2015}&=2015^{2015}\\ \left ( x^{2015} \right )^{x^{2015}}&=2015^{2015}\\ x^{2015}&=2015\\ x&=2015^{\frac{1}{2015}}\\ x&=\sqrt[2015]{2015} \end{aligned}.

 

 

Sumber Referensi

  1. Marwanta, Sigit Suprijanto, Herynugroho, Suwarni Murniati, Kamta Agus Sajak, Soetiyono. 2004. Matematika Interaktif 1A Kelas 1 SMA Semester Pertama. Jakarta: Yudistira.
  2. Sunardi, Slamet Waluyo, Sutrisno, Subagya. 2004. Matematika I A untuk SMA Kelas 1. Jakarta: PT Bumi Aksara.
  3. Susianto, Bambang. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: Grasindo.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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