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secondary 4 | A Maths
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Please help me thanku
Date Posted: 4 years ago
Views: 229
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done
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Assumption needed is p is positive except for 1.
Date Posted:
4 years ago
Thankuu!!
Shorter alternative :
2xp^(4logp(x)) = 486
xp^(logp(x⁴)) = 243
x(x⁴) = 243
x^5 = 3^5
x = 3
(Useful property to remember :
a^(loga(b)) = b
Similar : e^(ln b) = b
10^(lg b) = b
To see why it's so,
a^(loga(b)) = b
Take log a on both sides,
loga(b) = loga(b)
No need to assume p ≠ 1 . Because log base cannot be 1 or else it's undefined. Rather, it's deduced or inferred that p ≠ 1
2xp^(4logp(x)) = 486
xp^(logp(x⁴)) = 243
x(x⁴) = 243
x^5 = 3^5
x = 3
(Useful property to remember :
a^(loga(b)) = b
Similar : e^(ln b) = b
10^(lg b) = b
To see why it's so,
a^(loga(b)) = b
Take log a on both sides,
loga(b) = loga(b)
No need to assume p ≠ 1 . Because log base cannot be 1 or else it's undefined. Rather, it's deduced or inferred that p ≠ 1
This works because p^ and base p in the log are inverses of one another.
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