Question

Kuis:
[tex]\displaystyle\sqrt[(2-\log25)]{4^{log8}} +\:^{\sqrt{2}}\log\left(\log\frac{1}{0,01}\right)-\frac{\ln(\sqrt[3]{4})\ln(27)}{\ln(3)\ln(\sqrt{2})}=......[/tex]

Answer 1

Jawab:

[tex]\displaystyle\sf\sqrt[(2-\log\:25)]{4^{log\:8}}\:+ ^{\sqrt2}\log\left(\log\frac{1}{0,01}\right)-\frac{In(\sqrt[3]{4})In(27)}{In(3)In(\sqrt2)}\:=\:6[/tex]

Penjelasan dengan langkah-langkah:

Logaritma

Soal :

[tex]\displaystyle\sf\sqrt[(2-\log\:25)]{4^{log\:8}}\:+ ^{\sqrt2}\log\left(\log\frac{1}{0,01}\right)-\frac{In(\sqrt[3]{4})In(27)}{In(3)In(\sqrt2)} = \cdots[/tex]

Jawab :

Langkah 1 : Sederhanakan bentuk bentuknya

[tex]=\displaystyle\sf\sqrt[(2-\log\:25)]{4^{log\:8}}\:+ ^{\sqrt2}\log\left(\log\frac{1}{0,01}\right)-\frac{In(\sqrt[3]{4})In(27)}{In(3)In(\sqrt2)}[/tex]

• [tex]\displaystyle\sf\sqrt[(2-\log\:25)]{4^{log\:8}}\:\Rightarrow Gunakan\:sifat\:eksponen\:\sqrt[b]{x^a}\:=x^{\frac{a}{b}}[/tex]
• [tex]\displaystyle\sf \left(\log\frac{1}{0,01}\right)\Rightarrow \log\frac{1}{0,01}=\log\frac{1}{\frac{1}{100}}=\log100[/tex]
• [tex]\displaystyle\sf\frac{In(\sqrt[3]{4})In(27)}{In(3)In(\sqrt2)}\Rightarrow Gunakan\:sifat\:eksponen,sehingga =\frac{In(2^{\frac{2}{3}})In(3^3)}{In(3)In(2^{\frac{1}{2}})}[/tex]

Langkah 2 : Hitung !

[tex]= \displaystyle\sf4^{\frac{\log 8}{2-log25}}\:+ ^{\sqrt2}\log\left(\log\:100\right)-\frac{In(2^{\frac{2}{3}})In(3^3)}{In(3)In(2^{\frac{1}{2}})}[/tex]

[tex]= \displaystyle\sf2^{2\left(\frac{\log 8}{2-log25}\right)}\:+ ^{2^{\frac{1}{2}}}\log(2)-\frac{\frac{2}{3}\times3(In(2)In(3))}{\frac{1}{2}(In(2)In(3))}[/tex]

[tex]= \displaystyle\sf2^{2\left(\frac{\log 2^3}{2-log\:5^2}\right)}\:+2\times ^2\log(2)-\frac{\frac{2}{\cancel3}\times\cancel3\cancel{(In(2)In(3))}}{\frac{1}{2}\cancel{(In(2)In(3))}}[/tex]

[tex]= \displaystyle\sf2^{2\left(\frac{3\:\log 2}{2-2\:\log 5}\right)}\:+2\times 1-\frac{2}{\frac{1}{2}}[/tex]

[tex]= \displaystyle\sf2^{\cancel2\left(\frac{3\:\log 2}{\cancel2(1-\log5)}\right)}\:+2-4[/tex]

[tex]= \displaystyle\sf2^{\left(\frac{3\:\log 2}{(\log10-\log5)}\right)}\:-2\Rightarrow(\log a-\log b=\log{\frac{a}{b}})[/tex]

[tex]= \displaystyle\sf2^{\left(\frac{3\:\log 2}{\log2}\right)}\:-2[/tex]

[tex]= \displaystyle\sf2^3\:-2[/tex]

[tex]= \displaystyle\sf8\:-2[/tex]

[tex]= 6[/tex]

Kesimpulan :

Hasil dari  [tex]\displaystyle\sf\sqrt[(2-\log\:25)]{4^{log\:8}}\:+ ^{\sqrt2}\log\left(\log\frac{1}{0,01}\right)-\frac{In(\sqrt[3]{4})In(27)}{In(3)In(\sqrt2)}\:adalah\:6[/tex]

Pelajari lebih lanjut : Materi logaritma dan logaritma natural

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