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Soal dan Pembahasan Matematika Peminatan Kelas 10

Halo teman-teman..
Pada artikel kali ini akan mengupas soal matematika dan pembahasan peminatan kelas 10 semester 1 . Materi terkait diantaranya Eksponen dan Logaritma . Langsung saja yuk ke soalnya..

  1. Eksponen
    Tentukan himpunan penyelesaian dari persamaan eksponen ini \(9^{3x}-2^{3x+1}=27\) !
    Jawab :
    \(\begin{eqnarray}9^{3x}-2^{3x+1}&=&27\\3^{2(3x)}-2.3^{3x}.3^{1}-27&=&0\\(3^{3x})^{2}-6.3^{3x}-27&=&0\end{eqnarray}\)
    misal \(b=3^{3x}\), maka :
    \(\begin{eqnarray}b^{2}-6b-27&=&0\\(b-9)(b+3)&=&0\\b=9\hspace{1mm}atau\hspace{1mm}b&=&-3\end{eqnarray}\)
    untuk \(b=9\)
    \(\begin{eqnarray}b&=&9\\3^{3x}&=&9\\3^{3x}=3^{2}\\3x&=&2\\x&=&\frac{2}{3}\end{eqnarray}\)
    untuk \(b=-3\)
    \(\begin{eqnarray}3^{3x}&=&-3\\tidak\hspace{1mm}ada\hspace{1mm}nilai\hspace{1mm}x\hspace{1mm}yang\hspace{1mm}memenuhi\end{eqnarray}\)
    Sehingga HP={\(\frac{2}{3}\)}


  2. Eksponen
    Himpunan penyelesaian persamaan eksponen \(2^{2x^{2}+5x+3}=8^{3+2x}\) adalah ...
    Jawab :
    \(\begin{eqnarray}2^{2x^{2}+5x+3}&=&8^{3+2x}\\2^{2x^{2}+5x+3}&=&2^{3(3+2x)}\hspace{1mm}\rightarrow bilangan\hspace{1mm}pokok\hspace{1mm}sudah\hspace{1mm}sama,\hspace{1mm}maka\hspace{1mm}lihat\hspace{1mm}pangkatnya\\2x^{2}+5x-+3&=&3(3+2x)\\2x^{2}+5x+3&=&9+6x\\2x^{2}+5x+3-6x-9&=&0\\2x^{2}-x-6&=&0\\(2x+3)(x-2)&=&0\\x=-\frac{3}{2}\hspace{1mm}atau\hspace{1mm}x&=&2\end{eqnarray}\)
    HP = {\(-\frac{3}{2},2\)}


  3. Eksponen
    Tentukan nilai x yang memenuhi persamaan berikut :
    1. \(8^{x^{2}+x-6}=(\frac{2}{128})^{x-2}\)
    2. \(6^{x^{2}-1}=1\)
    3. \((\frac{1}{9})^{x^{2}-3x+2}=(\frac{1}{81})^{x^{2}-3x+2}\)
    4. \(7^{x^{2}-3x-2}=7^{-3x+1}\)
    5. \(11^{x^{2}+2x-5}=(\frac{1}{121})^{x}\)
    Jawab :

    1. \(\begin{eqnarray}8^{x^{2}+x-6}&=&(\frac{2}{128})^{x-2}\\8^{x^{2}+x-6}&=&(\frac{1}{64})^{x-2}\\8^{x^{2}+x-6}&=&8^{-2(x-2)}\hspace{1mm}\rightarrow bilangan\hspace{1mm}pokok\hspace{1mm}sudah\hspace{1mm}sama,\hspace{1mm}maka\hspace{1mm}lihat\hspace{1mm}pangkatnya\\x^{2}+x-6&=&-2(x-2)\\x^{2}+x-6&=&-2x+4\\x^{2}+x-6+2x-4&=&0\\x^{2}+3x-10&=&0\\(x+5)(x-2)&=&0\\x=-5\hspace{1mm}atau\hspace{1mm}x&=&2\end{eqnarray}\)
      HP = {\(-5,2\)}

    2. \(\begin{eqnarray}6^{x^{2}-1}&=&1\\6^{x^{2}-1}&=&6^{0}\\x^{2}-1&=&0\\(x-1)(x+1)&=&0\\x=1\hspace{1mm}atau\hspace{1mm}x&=&-1\end{eqnarray}\)
      HP = {\(-1,1\)}
       

    3. \(\begin{eqnarray}(\frac{1}{9})^{x^{2}-3x+2}&=&(\frac{1}{81})^{x^{2}-3x+2}\\(9)^{-1(x^{2}-3x+2)}&=&(9)^{-2(x^{2}-3x+2)}\\-1(x^{2}-3x+2)&=&-2(x^{2}-3x+2)\\-x^{2}+3x-2&=&-2x^{2}+6x-4\\-x^{2}+2x^{2}+3x-6x-2+4&=&0\\x^{2}-3x+2&=&0\\(x-1)(x-2)&=&0\\x=1\hspace{1mm}atau\hspace{1mm}x&=&2\end{eqnarray}\)
      HP = {\(1,2\)}

    4. \(\begin{eqnarray}7^{x^{2}-3x-2}&=&7^{-3x+1}\\x^{2}-3x-2&=&-3x+1\\x^{2}-3x+3x-2-1&=&0\\x^{2}-3&=&0\\(x-\sqrt{3})(x+\sqrt{3})&=&0\\x=\sqrt{3}\hspace{1mm}atau\hspace{1mm}x&=&-\sqrt{3}\end{eqnarray}\)
      HP = {\(-\sqrt{3},\sqrt{3}\)}

    5. \(\begin{eqnarray}11^{x^{2}+2x-5}&=&(\frac{1}{121})^{x}\\11^{x^{2}+2x-5}&=&11^{-2x}\\x^{2}+2x-5&=&-2x\\x^{2}+2x-5+2x&=&0\\x^{2}+4x-5&=&0\\(x+5)(x-1)&=&0\\x=-5\hspace{1mm}atau\hspace{1mm}x&=&1\end{eqnarray}\)
      HP = {\(-5,1\)}


  4. Eksponen
    Jumlah akar-akar dari persamaan eksponen \(3^{2x+1}-3^{x+3}-3^{x}=9\) adalah ...
    Jawab :
    \(\begin{eqnarray}3^{2x+1}-3^{x+3}-3^{x}&=&9\\3^{2x+1}-3^{x}.3^{3}-3^{x}-9&=&0\\3^{2x}.3^{1}-27.3^{x}-3^{x}-9&=&0\\3.(3^{x})^{2}-28.3^{x}-9&=&0\end{eqnarray}\)
    misal \(p=3^{x}\), maka :
    \(\begin{eqnarray}3p^{2}-28p-9&=&0\\(3p-1)(p-9)&=&0\\p=\frac{1}{3}\hspace{1mm}atau\hspace{1mm}p=9\end{eqnarray}\)
    Untuk \(p=\frac{1}{3}\) :
    \(\begin{eqnarray}3^{x}&=&\frac{1}{3}\\3^{x}&=&3^{-1}\\x&=&-1\end{eqnarray}\)
    Untuk \(p=9\) :
    \(\begin{eqnarray}3^{x}&=&9\\3^{x}&=&3^{2}\\x&=&2\end{eqnarray}\)
    HP = {\(-1,2\)}

  5. Logaritma
    Jika \(\alpha\) dan \(\beta\) adalah akar-akar dari persamaan logaritma \((^{27}log\frac{1}{x+1})^{2}=\frac{1}{9}\) , dengan \(\alpha>\beta\). Maka nilai dari \(\frac{\alpha}{\beta}\) adalah...
    Jawab :Ingat bahwa \(^{a}log\hspace{1mm}b=c\rightarrow b=a^{c}\)
    \(\begin{eqnarray}(^{27}log\frac{1}{x+1})^{2}&=&\frac{1}{9}\\(^{27}log\frac{1}{x+1})^{2}&=&\pm\sqrt{\frac{1}{9}}\\(^{27}log\frac{1}{x+1})^{2}&=&\pm\frac{1}{3}\end{eqnarray}\)
    Kondisi I :
    \(\begin{eqnarray}^{27}log\frac{1}{x+1}&=&\frac{1}{3}\\\frac{1}{x+1}&=&27^{\frac{1}{3}}\\\frac{1}{x+1}&=&3^{3(\frac{1}{3})}\\\frac{1}{x+1}&=&3^{1}\\1&=&3(x+1)\\1&=&3x+3\\3x&=&-2\\x&=&-\frac{2}{3}\end{eqnarray}\)
    Kondisi II :
    \(\begin{eqnarray}^{27}log\frac{1}{x+1}&=&-\frac{1}{3}\\\frac{1}{x+1}&=&27^{-\frac{1}{3}}\\\frac{1}{x+1}&=&3^{3(-\frac{1}{3})}\\\frac{1}{x+1}&=&3^{-1}\\\frac{1}{x+1}&=&\frac{1}{3}\\3&=&x+1\\x&=&2\end{eqnarray}\)
    HP = {\(-\frac{2}{3},2\)}


  6. Logaritma
    \(^{5}log\sqrt{27}.^{9}log\hspace{1mm}125+^{16}\hspace{1mm}log\hspace{1mm}32=...\)
    Jawab :
    \(\begin{eqnarray}^{5}log\sqrt{27}.^{9}log\hspace{1mm}125+^{16}\hspace{1mm}log\hspace{1mm}32&=&^{5}log\hspace{1mm}(3^{3})^{\frac{1}{2}}.^{3^{2}}log\hspace{1mm}5^{3}+^{2^{4}}\hspace{1mm}log\hspace{1mm}2^{5}\\&=&^{5}log\hspace{1mm}3^{\frac{3}{2}}.\frac{3}{2}.^{3}log\hspace{1mm}5+\frac{5}{4}.^{2}log\hspace{1mm}2\\&=&\frac{3}{2}.^{5}log\hspace{1mm}3.\frac{3}{2}.^{3}log\hspace{1mm}5+\frac{5}{4}.1\\&=&\frac{3}{2}.\frac{3}{2}.^{5}log\hspace{1mm}3.^{3}log\hspace{1mm}5+\frac{5}{4}\\&=&\frac{9}{4}.1+\frac{5}{4}\\&=&\frac{9}{4}+\frac{5}{4}\\&=&\frac{14}{4}\\&=&\frac{7}{2}\end{eqnarray}\)


  7. Logaritma
    Jika \(^{2}log\hspace{1mm}7=b\), maka \(^{8}log\hspace{1mm}\frac{1}{49}=...\)
    Jawab :
    \(\begin{eqnarray}^{8}log\hspace{1mm}\frac{1}{49}&=&^{2^{3}}log\hspace{1mm}7^{-2}\\&=&\frac{-2}{3}.^{2}log\hspace{1mm}7\\&=&-\frac{2}{3}.b\end{eqnarray}\)


  8. Logaritma
    Jika \(^{3}log\hspace{1mm}5=m\) dan \(^{2}log\hspace{1mm}3=n\), maka \(^{6}log\hspace{1mm}100=...\)
    Jawab :
    Ingat : \(^{a}log\hspace{1mm}b=\frac{^{p}log\hspace{1mm}b}{^{p}log\hspace{1mm}a},dengan\hspace{1mm}p>0\hspace{1mm}dan\hspace{1mm}p\neq1\)
    \(\begin{eqnarray}^{6}log\hspace{1mm}100&=&\frac{^{3}log\hspace{1mm}100}{^{3}log\hspace{1mm}6}\\&=&\frac{^{3}log\hspace{1mm}25+^{3}log\hspace{1mm}4}{^{3}log\hspace{1mm}3+^{3}log\hspace{1mm}2}\\&=&\frac{^{3}log\hspace{1mm}5^{2}+^{3}log\hspace{1mm}2^{2}}{1+^{3}log\hspace{1mm}2}\\&=&\frac{2.^{3}log\hspace{1mm}5+2.^{3}log\hspace{1mm}2}{1+\frac{1}{^{2}log\hspace{1mm}3}}\\&=&\frac{2m+\frac{2}{^{2}log\hspace{1mm}3}}{1+\frac{1}{n}}\\&=&\frac{2m+\frac{2}{n}}{1+\frac{1}{n}}\\&=&\frac{\frac{2mn+2}{n}}{\frac{n+1}{n}}\\&=&\frac{2mn+2}{n+1}\end{eqnarray}\)

Sampai sini dulu ya kita bahas soal matematika peminatan kelas 10 nya..
nantikan artikel dari ngerjainpr.com selanjutnya yaa..

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