²log 3 = x
³log 5 = y
[tex]_______________________[/tex]
buktikan jika hasil ⁵log 12 = (x+2)/(xy)
Jawaban:
Pembahasan
Sifat - sifat logaritma sebagai berikut :
[tex] \rm{}^{a}\log(a)=1 [/tex]
[tex] \rm{}^{a}\log(a^b)=b [/tex]
[tex] \rm{}^{a}\log(b)+{}^{a}\log(c)={}^{a}\log(bc) [/tex]
[tex] \displaystyle\rm{}^{a}\log(b)-{}^{a}\log(c)={}^{a}\log\left(\frac{b}{c}\right) [/tex]
[tex] \displaystyle\rm{}^{a}\log(b)=\frac{1}{{}^{b}\log(a)} [/tex]
[tex] \displaystyle\rm{}^{a}\log(b)=\frac{{}^{c}\log(b)}{{}^{c}\log(a)} [/tex]
[tex] \rm{}^{a}\log(b^d)=d \times {}^{a}\log(b) [/tex]
[tex] \displaystyle\rm{}^{a^c}\log(b)=\frac{1}{c} \times {}^{a}\log(b) [/tex]
[tex] \displaystyle\rm{}^{a^c}\log(b^d)={}^{a}\log(b^{\frac{d}{c}})=\frac{d}{c} \cdot {}^{a}\log(b) [/tex]
Diketahui bahwa ²log 3 = x dan ³log 5 = y
Buktikan jika hasil dari ⁵log 12 = [tex] \tt \frac{(x + 2)}{xy} [/tex]
[tex] {}^{5} log \: 12 = \frac{x + y}{xy} \\ [/tex]
[tex] \frac{ {}^{2}log \: 12 }{ {}^{2}log \: 5 } = \frac{x + 2}{xy} \\ [/tex]
[tex] \frac{ {}^{2} log \: (2 \times 6)}{ {}^{2} log \: 5} = \frac{x + 2}{xy} \\ [/tex]
[tex] \frac{ {}^{2} log \: 2 + {}^{2}log \: 6 }{ {}^{2}log \: 5 } = \frac{x + 2}{xy} \\ [/tex]
[tex] \frac{1 + {}^{2}log \: (2 \times 3) }{ {}^{2}log \: 5 } = \frac{x + 2}{xy} \\ [/tex]
[tex] \frac{1 + {}^{2}log \: 2 + {}^{2}log \: 3 }{ {}^{2}log \: 5 } = \frac{x + 2}{xy} \\ [/tex]
[tex] \frac{1 + 1 + {}^{2} log \: 3}{ {}^{2} log \: 5} = \frac{x + 2}{xy} \\ [/tex]
[tex] \frac{2 + {}^{2}log \: 3 }{ {}^{2} log \: 5} = \frac{x + 2}{xy} \\ [/tex]
Ingat bahwa; ²log 3 = x dan ³log 5 = y
Disini bisa kita ubah satu persatu;
²log 3 = ³log 2 = [tex] \tt \frac{1}{x} [/tex]
³log 5 = ⁵log 3 = [tex] \tt \frac{1}{y} [/tex]
²log 3 . ³log 5 = ²log 5 = xy
Maka:
[tex] \frac{2 + {}^{2} log \: 3 }{ {}^{2} log \: 5} = \frac{x + 2}{xy} \\ [/tex]
[tex] \frac{2 + x}{xy} = \frac{x + 2}{xy} \\ [/tex]
atau
[tex] \huge \frac{x + 2}{xy} = \frac{x + 2}{xy} \to \: TERBUKTI[/tex]
Terbukti benar bahwa:
[tex]{}^5\log12=\dfrac{x+2}{xy}[/tex]
Pembahasan
Logaritma
Diketahui:
- [tex]{}^2\log3=x[/tex]
- [tex]{}^3\log5=y[/tex]
Bentuk-bentuk lain yang ekuivalen berdasarkan sifat-sifat logaritma antara lain adalah:
[tex]\begin{aligned}&{}^2\log3=x\\&\Rightarrow\frac{{}^5\log3}{{}^5\log2}=x\quad...(i)\\&\Rightarrow{}^5\log2=\frac{{}^5\log3}{x}\quad...(ii)\\&.........................................\\&{}^3\log5=y\\&\Rightarrow{}^5\log3=\frac{1}{y}\quad...(iii)\\&\!\!\stackrel{(ii)}{\implies}{}^5\log2=\frac{1}{xy}\quad...(iv)\\&.........................................\\\end{aligned}[/tex]
Kemudian, akan dibuktikan bahwa:
[tex]\large\text{$\begin{aligned}\boxed{\ {}^5\log12=\frac{x+2}{xy}\ }\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&\textsf{Ruas kiri}={}^5\log12\\&{=\ }{}^5\log\left(2^2\cdot3\right)\\&{=\ }{}^5\log\left(2^2\right)+{}^5\log3\\&{=\ }2\cdot{}^5\log2+{}^5\log3\\&\ ...\ \textsf{substitusi dari $(iii)$ dan $(iv)$}\\&{=\ }2\cdot\frac{1}{xy}+\frac{1}{y}\\&{=\ }\frac{2}{xy}+\frac{x}{xy}\\&{=\ }\frac{x+2}{xy}=\textsf{Ruas kanan}\\&\blacksquare\quad\textsf{TERBUKTI}\end{aligned}$}[/tex]
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