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secondary 4 | A Maths
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Looking at -log8 x^3, I really don't know how it became -log4 x^2. Pls show the method. Thank you!
Date Posted: 4 years ago
Views: 380
log8(x³)
= log4(x³)/log4(8)
= 3log4(x)/log4(4^(3/2))
= 3log4(x)/(3/2log4(4))
= 3log4(x)/(3/2)
= 2log4(x)
= log4(x²)
= log4(x³)/log4(8)
= 3log4(x)/log4(4^(3/2))
= 3log4(x)/(3/2log4(4))
= 3log4(x)/(3/2)
= 2log4(x)
= log4(x²)
Or
log8(x³)
= log4(x³)/log4(8)
= 3log4(x)/log4(4 x 2)
= 3log4(x)/(log4(4) + log4(2))
= 3log4(x)/(log4(4) + log4(4^½) )
= 3log4(x)/(1 + ½)
= 3log4(x)/(3/2)
= 2log4(x)
= log4(x²)
log8(x³)
= log4(x³)/log4(8)
= 3log4(x)/log4(4 x 2)
= 3log4(x)/(log4(4) + log4(2))
= 3log4(x)/(log4(4) + log4(4^½) )
= 3log4(x)/(1 + ½)
= 3log4(x)/(3/2)
= 2log4(x)
= log4(x²)
See 1 Answer
done
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It’s sad that the book skips crucial steps at times.
Date Posted:
4 years ago
answer schemes tend to be more brief. Just showing the crucial steps. The manipulation is left to the student
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