J's answer to QN's Junior College 1 H1 Maths Singapore question.
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2 ln 2x = 1 + ln a, where a > 0
ln (2x)² = ln e + ln a
ln 4x² = ln ae
4x² = ae
x² = ae/4
x = √ae / 2
ln (2x)² = ln e + ln a
ln 4x² = ln ae
4x² = ae
x² = ae/4
x = √ae / 2
Date Posted:
4 years ago
Note that the negative result is rejected
i.e x ≠ -√ae / 2 because that would mean:
2 ln 2x = 2 ln 2(-√ae / 2 )
= 2 ln (-√ae)
And since a > 0, -√ae < 0 and thus 2 ln (-√ae) is undefined.
i.e x ≠ -√ae / 2 because that would mean:
2 ln 2x = 2 ln 2(-√ae / 2 )
= 2 ln (-√ae)
And since a > 0, -√ae < 0 and thus 2 ln (-√ae) is undefined.
Alternative working :
2 ln 2x = 1 + ln a
2 ln 2x = ln e + ln a
2 ln 2x = ln ae
ln 2x = ½ ln ae
ln 2x = ln (ae)¹/² = ln √ae
2x = √ae
x = √ae / 2
2 ln 2x = 1 + ln a
2 ln 2x = ln e + ln a
2 ln 2x = ln ae
ln 2x = ½ ln ae
ln 2x = ln (ae)¹/² = ln √ae
2x = √ae
x = √ae / 2
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