1 2log 4+2log82 2log 20-2log534log32.... Question from @rinanatalia872

August 2018 2 38 Report

1 2log 4+2log8
2 2log 20-2log5
34log32

ayuningtyasrb $^2log 4 + {^2}log 8 = 2 + 3 = 5$

$^2log 20 - {^2}log 5 \\ = {^2}log\frac{20}{5} \\ = {^2}log 4 \\ = 2$

$^4log 32 \\ = {^{2^2}}log 2^5 \\ = \frac{5}{2}.{^2}log 2 \\ = \frac{5}{2}$

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IcukSugiarto $\bold{\underline{Solution\ 1 :}}\\~^{2}log\ 4 +~^{2}log\ 8\\=~^{2}log\ (4 \times 8)\\=~^{2}log\ 32\\=~^{2}log\ 2^{5}\\=5~^{2}log\ 2\\=5(1)\\=5$

$\bold{\underline{Solution\ 2 :}}\\~^{2}log\ 20-~^{2}log\ 5\\=~^{2}log\ \frac{20}{5}\\=~^{2}log\ 4\\=~^{2}log\ 2^{2}\\=2~^{2}log\ 2\\=2(1)\\=2$

$\bold{\underline{Solution\ 3 :}}\\~^{4}log\ 32\\=~^{2^{2}}log\ 2^{5}\\= \frac{5}{2} ~^{2}log\ 2\\= \frac{5}{2} (1)\\= ~~\frac{5}{2}$

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