Matematika [4/10][tex]~[/tex][tex] \rm {6}^{a + b - 1} = {3}^{a} \times {2}^{b + 1} [/tex]a) a = ²log 5 dan b = ³log 4b) a = ²log 3 dan b = ³log 4c) a = ³log 4 dan b = ²log 3d) a = ²log 3 dan b = ³log 4Jawabnya pakai penalaran:)​

brainroweulv – June 09, 2023

[4/10]
[tex]~[/tex]
[tex] \rm {6}^{a + b - 1} = {3}^{a} \times {2}^{b + 1} [/tex]
a) a = ²log 5 dan b = ³log 4
b) a = ²log 3 dan b = ³log 4
c) a = ³log 4 dan b = ²log 3
d) a = ²log 3 dan b = ³log 4


Jawabnya pakai penalaran:)

​

Jawaban:

b)

Penjelasan dengan langkah-langkah:

[tex] {6}^{a + b - 1} = {3}^{a} \times {2}^{b + 1} \\ {6}^{a} \times {6}^{b} \times {6}^{ - 1} = {3}^{a} \times {2}^{b} \times 2 \\ {2}^{a} \times {3}^{a} \times {2}^{b} \times {3}^{b} \times \frac{1}{6} = {3}^{a} \times {2}^{b} \times 2 \\ {2}^{a} \times {3}^{b} \times \frac{1}{6} = 2 \\ {2}^{a} \times {3}^{b} = 12 \\ {2}^{ {}^{2} log_{3}} \times {3}^{ {}^{3} log_{4}} = 12 \\ 3 \times 4 = 12 \: {terbukti}[/tex]

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