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Let Solve This MTH. by Seuncoded (m): 10:59pm on April 15
8. Given that Log₄(y+1) + Log₄(x/2) = z and Log₂(y-1) - Log₂ x = z-1, show that y²=1 + 8^z and find the possible values for x and y if z=1
Solution

Log₄(y+1)+Log₄(x/2)=z

Log₄[(y+1)(x/2)] = z

4^z = (y+1)(x/2)

(4^z) × 2 = (y+1)(x)

x = (2×4^z)/(y+1) - -- (1)


Log₂(y-1) - Log₂ x = z-1

Log₂[(y-1)/x] = z-1

2^(z-1) = (y-1)/x

x × 2^(z-1) = (y-1)

x = (y-1)/[2^(z-1)] - -- (2)

Equating (1) and (2)

(2×4^z)/(y+1)=(y-1)[2^(z-1)]

2¹×2^(2z)×2^(z-1)=(y-1)(y+1)

2^(1+2z) × 2^(z-1)=y²-1

2^(1+2z+z-1) = y² - 1

2^(3z) = y² - 1

8^z = y² - 1

Hence, y² = 1 + 8^z

Lets now find the possible values for x and y if z=1

From, y² = 1 + 8^z

Substitute z=1

We have, y² = 1 + 8

y² = 9

y = ± √9

y = ±3

Recall that, x = (y-1)/[2^(z-1)]

Substitute z=1

x = y - 1

When y = + 3
x = 3 - 1
x = 2

When y = - 3
x = - 3 - 1
x = - 4



9. The rth term Ur of a Sequence U₁, U₂,.... is given by Ur = U(r—₁ ) + 8(r-1). If U₂=9, write down the values of U₁, U₃, U₄ and obtain the express for Ur in terms of r.

NB: In U(r—₁ ), the (r—₁) is Subscript

Solution
Ur = U(r—₁ ) + 8(r-1)

Given that U₂ = 9

U₂ = U(₂—₁) + 8(2-1)
U₂ = U₁ + 8
U₁ + 8 = 9
U₁ = 1

U₃ = U(₃—₁) + 8(3-1)
U₃ = U₂ + 8(2)
U₃ = 9 + 16
U₃ = 25

U₄ = U(₄—₁) + 8(4-1)
U₄ = U₃ + 8(3)
U₄ = 25 + 24
U₄ = 49

Hence,
U₁ = 1
U₃ = 25
U₄ = 49

Now let's obtain the expression for Ur in terms of r

We obtain the expression for Ur from the sequence:

1, 9 , 25, 49
/ / /
/ / /
8 16 24

Since the common difference at the second is equal, the sequence is a Quadratic sequence

The rth term of a quadratic sequence is of the form

Ur = ar² + br + c

For the Sequence:
1, 9, 25, 49

U₁ = a(1²) + b(1) + c =1

a + b + c = 1 - - - - (1)

U₂ = a(2²) + b(2) + c=9

4a + 2b + c = 9 - - - - (2)

U₃ = a(3²) + b(3) + c=25

9a + 3b + c = 25 - - - (3)
U₄ =a(4²) + b(4) + c =49

16a + 4b + c = 49 - - (4)

Solving for a, b & c picking 3 equations only

a = 4
b = - 4
c = 1

Substitute into Ur = ar² + br + c

Ur = 4n² - 4n + 1



10. The first, third and ninth terms of a linear sequence (A.P) are the first three terms of an exponential sequence (G.P). If the seventh term of the linear sequence is 14. Calculate:
(i) The twentieth term of the linear sequence
(ii) The sum of the first twelve terms of the exponential sequence.

Solution.
The 1st, 3rd & 9th of the A.p are

T₁ = a
T₃ = a + 2d
T9 = a + 8d

But, T7 = a + 6d = 14
==> a = 14 - 6d

Substitute a into T₁, T₃ & T9

T₁ = a = 14 - 6d

T₃ = a + 2d
=14 - 6d +2d
=14 - 4d

T9 = a + 8d
= 14 - 6d + 8d
= 14 + 2d

Since they are equal to the first three terms of a G.p it implies that;

14 - 6d = a
14 - 4d = ar
14 + 2d = ar²

Recall that, to find the common ratio of a G.p

T₂/T₁ = T₃/T₂

14 - 4d 14 + 2d
----------- = - - - - - - -
14 - 6d 14 - 4d

Cross multiply

(14-4d)(14-4d) = (14+2d)(14-6d)

196 - 56d - 56d + 16d² = 196 - 84d + 28d - 12d²

Re-arrange

16d² - 112d + 196 = - 12d² - 56d + 196

Collect like terms

28d² - 56d = 0

d = 2 or 0

From, a = 14 - 6d

a = 14 - 6(2)

a = 2

From, r = (14-4d)/(14-6d)

r = [14 - 4(2)]/[14 - 6(2)]

r = 6/2

r = 3

Thus,
a = 2
d = 2
r = 3

(i) T20 = a + 19d
= 2 + 19(2)
= 2 + 38
= 40

(ii) Sn = [a(rⁿ-1)]/(r-1)

S₁₂ = 2(3¹²-1)/(3-1)

S₁₂ = 2(3¹²-1)/2

S₁₂ = 3¹² - 1

S₁₂ = 531,440
Let Solve This MTH - Academic Tips

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Re: Let Solve This MTH. by Seuncoded (m): 10:59pm on April 15
8. Given that Log₄(y+1) + Log₄(x/2) = z and Log₂(y-1) - Log₂ x = z-1, show that y²=1 + 8^z and find the possible values for x and y if z=1
Solution

Log₄(y+1)+Log₄(x/2)=z

Log₄[(y+1)(x/2)] = z

4^z = (y+1)(x/2)

(4^z) × 2 = (y+1)(x)

x = (2×4^z)/(y+1) - -- (1)


Log₂(y-1) - Log₂ x = z-1

Log₂[(y-1)/x] = z-1

2^(z-1) = (y-1)/x

x × 2^(z-1) = (y-1)

x = (y-1)/[2^(z-1)] - -- (2)

Equating (1) and (2)

(2×4^z)/(y+1)=(y-1)[2^(z-1)]

2¹×2^(2z)×2^(z-1)=(y-1)(y+1)

2^(1+2z) × 2^(z-1)=y²-1

2^(1+2z+z-1) = y² - 1

2^(3z) = y² - 1

8^z = y² - 1

Hence, y² = 1 + 8^z

Lets now find the possible values for x and y if z=1

From, y² = 1 + 8^z

Substitute z=1

We have, y² = 1 + 8

y² = 9

y = ± √9

y = ±3

Recall that, x = (y-1)/[2^(z-1)]

Substitute z=1

x = y - 1

When y = + 3
x = 3 - 1
x = 2

When y = - 3
x = - 3 - 1
x = - 4



9. The rth term Ur of a Sequence U₁, U₂,.... is given by Ur = U(r—₁ ) + 8(r-1). If U₂=9, write down the values of U₁, U₃, U₄ and obtain the express for Ur in terms of r.

NB: In U(r—₁ ), the (r—₁) is Subscript

Solution
Ur = U(r—₁ ) + 8(r-1)

Given that U₂ = 9

U₂ = U(₂—₁) + 8(2-1)
U₂ = U₁ + 8
U₁ + 8 = 9
U₁ = 1

U₃ = U(₃—₁) + 8(3-1)
U₃ = U₂ + 8(2)
U₃ = 9 + 16
U₃ = 25

U₄ = U(₄—₁) + 8(4-1)
U₄ = U₃ + 8(3)
U₄ = 25 + 24
U₄ = 49

Hence,
U₁ = 1
U₃ = 25
U₄ = 49

Now let's obtain the expression for Ur in terms of r

We obtain the expression for Ur from the sequence:

1, 9 , 25, 49
/ / /
/ / /
8 16 24

Since the common difference at the second is equal, the sequence is a Quadratic sequence

The rth term of a quadratic sequence is of the form

Ur = ar² + br + c

For the Sequence:
1, 9, 25, 49

U₁ = a(1²) + b(1) + c =1

a + b + c = 1 - - - - (1)

U₂ = a(2²) + b(2) + c=9

4a + 2b + c = 9 - - - - (2)

U₃ = a(3²) + b(3) + c=25

9a + 3b + c = 25 - - - (3)
U₄ =a(4²) + b(4) + c =49

16a + 4b + c = 49 - - (4)

Solving for a, b & c picking 3 equations only

a = 4
b = - 4
c = 1

Substitute into Ur = ar² + br + c

Ur = 4n² - 4n + 1



10. The first, third and ninth terms of a linear sequence (A.P) are the first three terms of an exponential sequence (G.P). If the seventh term of the linear sequence is 14. Calculate:
(i) The twentieth term of the linear sequence
(ii) The sum of the first twelve terms of the exponential sequence.

Solution.
The 1st, 3rd & 9th of the A.p are

T₁ = a
T₃ = a + 2d
T9 = a + 8d

But, T7 = a + 6d = 14
==> a = 14 - 6d

Substitute a into T₁, T₃ & T9

T₁ = a = 14 - 6d

T₃ = a + 2d
=14 - 6d +2d
=14 - 4d

T9 = a + 8d
= 14 - 6d + 8d
= 14 + 2d

Since they are equal to the first three terms of a G.p it implies that;

14 - 6d = a
14 - 4d = ar
14 + 2d = ar²

Recall that, to find the common ratio of a G.p

T₂/T₁ = T₃/T₂

14 - 4d 14 + 2d
----------- = - - - - - - -
14 - 6d 14 - 4d

Cross multiply

(14-4d)(14-4d) = (14+2d)(14-6d)

196 - 56d - 56d + 16d² = 196 - 84d + 28d - 12d²

Re-arrange

16d² - 112d + 196 = - 12d² - 56d + 196

Collect like terms

28d² - 56d = 0

d = 2 or 0

From, a = 14 - 6d

a = 14 - 6(2)

a = 2

From, r = (14-4d)/(14-6d)

r = [14 - 4(2)]/[14 - 6(2)]

r = 6/2

r = 3

Thus,
a = 2
d = 2
r = 3

(i) T20 = a + 19d
= 2 + 19(2)
= 2 + 38
= 40

(ii) Sn = [a(rⁿ-1)]/(r-1)

S₁₂ = 2(3¹²-1)/(3-1)

S₁₂ = 2(3¹²-1)/2

S₁₂ = 3¹² - 1

S₁₂ = 531,440
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