Pembahasan soal logaritama (1)


Untuk soal nomor 1.
Tentukan nilai dari x^{\log y-\log z}.y^{\log z - \log x}.z^{\log x - \log y}
Jawab

misal a = x^{\log y-\log z}.y^{\log z - \log x}.z^{\log x - \log y}, ruas kiri dan ruas kanan dijadikan log sehingga:
\log a = \log (x^{\log y-\log z}.y^{\log z - \log x}.z^{\log x - \log y})
dengan menggunakan sifat logaritma kita diuraikan menjadi:
\log a = \log (x)^{\log y-\log z}+\log(y)^{\log z - \log x}
+ \log (z)^{\log x - \log y}

\log a = (\log y-\log z)\log x +(\log z - \log x) \log y +(\log x - \log y)\log z

\log a = \log y\log x-\log z\log x +\log y\log z - \log x \log y
+\log x\log z - \log y\log z
\log a = 0
sehingga a = 1
Jadi nilai dari x^{\log y-\log z}.y^{\log z - \log x}.z^{\log x - \log y}= 1

Untuk soal nomor 2

Tentukan nilai x, {}^{2}\log x+{}^{4}\log x+{}^{16}\log x = \frac{21}{4}.

Jawab.

{}^{2}\log x+{}^{4}\log x+{}^{16}\log x = \frac{21}{4} .
{}^{2}\log x+{}^{2^2}\log x+{}^{2^4}\log x = \frac{21}{4}
{}^{2}\log x+(\frac{1}{2}){}^{2}\log x+(\frac{1}{4}){}^{2}\log x = \frac{21}{4}
(1+\frac{1}{2}+\frac{1}{4}).{}^{2}\log x = \frac{21}{4}
(\frac{7}{4}).{}^{2}\log x = \frac{21}{4}
{}^{2}\log x = \frac{21}{4}.(\frac{4}{7})
{}^{2}\log x = 3
x = 2^3=8
Jadi nilai x = 8 memenuhi persamaan {}^{2}\log x+{}^{4}\log x+{}^{16}\log x = \frac{21}{4}

Untuk soal nomor 3
Berapakah nilai dari \frac{1}{{}^{a}\log (abc)}+\frac{1}{{}^{b}\log (abc)}+\frac{1}{{}^{c}\log (abc)} .

Jawab.

ingat sifat logaritma {}^{a}\log b = \frac{1}{{}^{b}\log a}, sehingga :
\frac{1}{{}^{a}\log (abc)}+\frac{1}{{}^{b}\log (abc)}+\frac{1}{{}^{c}\log (abc)}

={}^{(abc)}\log a +{}^{(abc)}\log b +{}^{(abc)}\log c

= {}^{(abc)}\log (a.b.c)
= 1.
Jadi nilai dari \frac{1}{{}^{a}\log (abc)}+\frac{1}{{}^{b}\log (abc)}+\frac{1}{{}^{c}\log (abc)} = 1
tex

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