logarithms....plzzz ans...thnk u
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3(log5-log3) - (log5-2log6) = 2-log n
We can write 2 as log 100 because
log 100 = log^100 base 10 = log^10² base 10= 2(1)=2
2log6 = log6²=log36
3log5/3 - log 5/36 + log n = log100
log (5/3)³ - log5/36+log n = log 100
log 125/27-log5/36+log n = log 100
log 125×36/27×5 + log n = log 100
log 100/3 + log n = log 100
log 100×n/3 = log 100
log 100n/3 = log100
Remove logarithms on both sides,
100n/3 = 100
n = 100×3/100
n = 3
Hope it helps
We can write 2 as log 100 because
log 100 = log^100 base 10 = log^10² base 10= 2(1)=2
2log6 = log6²=log36
3log5/3 - log 5/36 + log n = log100
log (5/3)³ - log5/36+log n = log 100
log 125/27-log5/36+log n = log 100
log 125×36/27×5 + log n = log 100
log 100/3 + log n = log 100
log 100×n/3 = log 100
log 100n/3 = log100
Remove logarithms on both sides,
100n/3 = 100
n = 100×3/100
n = 3
Hope it helps
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