Solve this NECO past questions 2016/2017

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Geaty
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PostGeaty on 28th June 2016, 3:05 pm

Q7a) Find d value of n if
32n+1 - 4(3n+1) + 9 = 0

7b) Solve d eqn:
log(x2+4) = 2 + logx - log20

Q9a) A two-digit number is such that its tens digit is greater than its unit digit by 5. If d number is 14 less than 3 times d product of its digits, find d numbers
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PostAdmin on 28th June 2016, 3:21 pm

Log(x2 + 4) = 2 + logx - log20
log(x2 + 4) = log100 + logx - log20
log(x2 + 4) = log (100 X x)
                              20

Now, you cross multiply
given you 20(x2 + 4) = 100 X x
20x2 + 80 = 100x

now divid by 20
20x2/ - 100x + 80 = 0
20         20      20


x2 - 5x + 4 = 0
x2 - 4x - x + 4 = 0
(x2 - 4x)(x + 4)
x(x - 4) -1(x - 4)
(x - 1)(x - 4)
x = 1 or 4


Last edited by Paradise on 29th June 2016, 9:22 am; edited 1 time in total


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Geaty
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PostGeaty on 28th June 2016, 4:25 pm

32n+1 - 4(3n+1) + 9 = 0

Note that
32n+1 = 31.32n = 31.(3n)2

while
4(3n+1) = 4(31.3n)

Now
3(3n)2 - 4 x 3(3n) + 9 = 0
3(3n)2 - 12(3n) + 9 = 0
Let us call 3n = y

therefore 3y2 - 12y + 9 = 0

Divide by 3
3yn - 12y + = 0
3       3       3

we will have
y2 - 4y + 3 = 0
y2 - 3y - y + 3 = 0
(y2 - 3y)(-y +3)
y(y - 3) - 1(y - 3)
(y - 1)(y - 3)
y=1 or 3

therefore when y=1
3n = y
n = 0

when y = 3
3n = y
n = 1

n= 0 or 1    Answer
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PostWimpy on 29th June 2016, 11:14 am

Very cool


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