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Question
secondary 4 | E Maths
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I managed to solve 10b and 11 but only with A Math method (ie substitution and log). But this is an E Math question - does anyone have any idea how to approach this that's within the E Math syllabus?
There might have have been some typo or mis-pasting of questions by the setter/teacher.
If 10b) were to be solved using E Math methods only then there's only 1 possible answer (no using of logarithms)
And it's by observation. (actually guess and check)
I will use superscript n in place of x.
(superscript x not available on my keyboard)
3²ⁿ+¹ + 6 = 3ⁿ+²
(3²ⁿ)(3¹) + 6 = (3ⁿ)(3²)
3(3ⁿ)² + 6 = 9(3ⁿ)
(3ⁿ)² + 2 = 3(3ⁿ)
By observation,
(3°)² + 2
= 1² + 2
= 1 + 2
= 3
= 3 x 1
= 3(3°)
So n = 0
Or,
6
= 9 - 3
= 3² - 3¹
= 3°+² - 3²(°)+¹
By oberservation, n = 0
Or
(3ⁿ)² + 2 = 3(3ⁿ)
2 = 3(3ⁿ) - (3ⁿ)²
But 2 = 3 - 1
= 3 x 1 - 1²
= 3(3°) - (3°)²
By observation, n = 0
See 1 Answer
In order to be three multiples
X got to be zero