O lvl A Maths: Indices and Logarithms

Given that log_p(x²y) = 8 and log_p(y² /x) = 6, evaluate

(i) log_p(xy)
(ii) log_p(y / x)

if y/x = 9, find the value of p, of x and of y.

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Answer:

log_p(x²y) = 8
===> log_px² + log_py = 8
===> 2log_px + log_py = 8 ------------------------- (1)

log_p(y² /x) = 6
===> log_py² - log_p x = 6
===> 2log_py - log_p x = 6 ------------------------- (2)

(1) + 2 * (2):
log_py + 4 log_py = 8 + 2(6)
5 log_py = 2o
log_py = 4 ------------------------------------------------------- (3)

Sub (3) into (1)
2 log_px + 4 = 8
log_px = 2 ------------------------------------------------------- (4)

(i) log_p(xy)
= log_px + log_py
= 4 + 2
= 6

(ii) log_p(y/x)
= log_py - log_px
= 4 - 2
= 2


If y/x = 9, from (ii)
log_p9 = 2
p² = 9
p = 3 or -3 (rej for log to be defined)
Thus, p = 3.

Thus, log_px = 2
x = p²
x = 9

log_py = 4
y = p⁴
y = 81

O lvl A Maths: Indices and Logarithms

Solve the equation log₃ 2 + log₃(x + 4) = 2 log₃ x.

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Answer:

log₃ 2 + log₃(x + 4) = 2 log₃ x
log₃ 2 (x + 4) = log₃ x²
2 (x + 4) = x²
x² - 2x - 8 = 0
(x - 4)(x + 2) = 0
x = 4 or -2

For log₃ x to be defined, x = 4.

Note: x = -2 is rejected because log₃ x will be undefined.

O lvl A Maths: Indices and Logarithms

(i) By using the substitution y = 2^x, find the value of x such that

4^x+1 = 2 - 7(2^x).


(ii) Express 3^x(2^2x) = 7(5^x) in the form a^x = b. Hence find x.

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Answer:

(i) Since 4^x+1 = 4(4^x) = 4(2²)^x = 4(2^x)², the equation becomes

4(2^x)² = 2 - 7(2^x)

Substituting 2^x = y,
4y² = 2 - 7y
4y² +7y - 2 = -
(4y - 1)(y + 2) = 0
y = 0.25 or y = -2
2^x = 2^-2 or 2^x = -2 (N.A. as no solution)
Hence, x = -2


(ii) Since 3^x(2^2x) = 3^x(2²)^x = 12^x,
12^x = 7(5^x)
(12 / 5)^x = 7

Taking log on both sides:

x lg (12 / 5) = lg 7
x = lg 7 / lg (12/5)
x = 2.22

O lvl A Maths: Indices and Logarithms

(a) Solve for x in the equation
(2x)^lg2 = (7x)^lg7

(b) Solve (log_aX)^log_bX = X where a, b are positive real numbers except 1, leave the answer in terms of a and b.

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Answer:

(a)
(2x)^{lg 2} = (7x)^lg7
(2^{lg 2})(x^{lg 2})=(7^{lg 7})(x^{lg 7})
(x^{lg 2}) / (x^{lg 7})=(7^{lg 7}) / (2^{lg 2})
x^{(lg2 - lg7)} = (7^{lg 7}) / (2^{lg 2})

Bring over the power to the other side of the equation
x = [(7^{lg 7}) /(2^{lg 2})]^{{1/(lg2-lg7)}}


(b)
(log_a x)^{log_b x} = x

Let x = b^{log_b x}
(loga x)^{log_b x} = b^{log_b x}

Same power, equate base
log_a x = b
x = a^b

O lvl A Maths: Indices and Logarithms

From http://www.sgforums.com/forums/2297/topics/332604 courtesy of Ahm97sic

(1) Evaluate log_n 9 / log_n 4

(2) Given that zy = 2^{[log₂ (z + x)]}, show that z = x / (y-1)

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Answer:

(1)
log_n 9 / log_n 4
= (lg 9 / lg n) / (lg 4 / lg n)
= lg 9 / lg 4
= (2 * lg 3) / (2 * lg 2)
= lg 3 / lg 2
= 1.585

(2)
zy = 2^{[log₂ (z + x)]}
zy = z + x ===> 2^{[log₂ x]} = x, or n^{[log_n x]} = x
zy - z = x
z(y - 1) = x
z = x/(y - 1) (shown)

O lvl A Maths: Indices and Logarithms

Show that 5ⁿ + 5ⁿ⁺¹ + 5ⁿ⁺² is divisible by 31 for all positive integer values of n

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Answer:

5ⁿ + 5ⁿ⁺¹ + 5ⁿ⁺²
= 5ⁿ + 5ⁿ 5¹ + 5ⁿ 5²
= 5ⁿ ( 1 + 5¹ + 5² )
= 5ⁿ (31)

Hence, 5ⁿ + 5ⁿ⁺¹ + 5ⁿ⁺² is divisible by 31 for all positive integer values of n (shown).

O lvl A Maths: Indices and Logarithms

(a) x^3/2 - 8x^-3/2 = 7

(b) e^2x - 5e^-2x = 4

(c) 2^x+3 = 9

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Answer:

(a)


Substitute y = -1 into y = x^3/2
-1 = x^3/2 (rejected)


Substitute y = -1 into y = x^3/2
8 = x^3/2
2³ = x^3/2
x = 2²
x = 4 (ans)


(b)


Substitute y = -1 into y = e^2x
-1 = e^2x (rejected)


Substitute y = 5 into y = e^2x
5 = e^2x
ln 5 = ln e^2x
1.61 = 2x ln e
2x = 1.61
x = 0.805 (ans)


(c)

O lvl A Maths: Indices and Logarithms

(a) Given that , evaluate x, y and z.

(b) Given that 2^2x+2 * 5^x-1 = 8^x * 5^2x, evaluate 10^x.

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Answer:
(a)








By comparison,
x = 10 1/2
y = 7 1/2
z = 1 1/2 (ans)

(b)
2^2x+2 * 5^x-1 = 8^x * 5^2x

2^2x2² * 5^x5^-1 = (2³)^x * 5^2x
4/5 = 2^3x-2x * 5^2x-x
4/5 = 2^x * 5^x
10^x = 4/5 (ans)

O lvl A Maths: Indices and Logarithms

(a) Find, in its simplest form, the product of a^⅓ + b^⅔ and a^⅔ - a^⅓ b^⅔ + b^4/3.


(b) Simplify

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Answer:

(a)




(b)

O lvl A Maths: Indices and Logarithms

Question from http://www.sgforums.com/forums/2297/topics/322475

1) If a^2x-1 = b^1-3y and a^3x-1 = b^2y-2, show that 13xy = 7x + 5y - 3.

2) Given that log₅ x = 4 log_x 5, calculate the possible values of x.

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Answer:

1)

Log to base 10 on the equations

lg (a^2x-1) = lg (b^1-3y)
(2x - 1) lg a = (1 - 3y) lg b
(2x - 1) /(1 - 3y) = lg b / lg a----(1)

lg (a^3x-1) = lg (b^2y-2)
(3x - 1) lg a = (2y - 2) lg b
(3x-1) /(2y-2) = lg b / lg a----(2)

(1) = (2)

thus
(2x - 1) /(1 - 3y) = (3x-1) /(2y-2)
4xy + 2 - 4x- 2y = 3x + 3y- 1- 9xy

Hence 13xy = 5y + 7x - 3

2) log₅ x = 4 log_x 5
log₅ x = 4 log₅ 5 / log₅ x

[log₅ x]² = 4

log₅ x = ±2

x=5^-2 or 5²

x= ¹/₂₅ or 25

O lvl A Maths: Indices and Logarithms

Question from http://www.sgforums.com/forums/2297/topics/309850

Given that log_b(xy²) = m and log_b (x³y)= n, express log_b y/x and log_b √(xy) in terms of m and n.

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Answer:


(1): log_b (xy²) =m
(2): log_b (x³y)=n


(1) becomes log_b x + 2 log_b y = m
(2) becomes 3 log_b x + log_b y = n

2 * (2) – (1) : 5 log_b x = 2n – m

(3): log_b x = 0.2 (2n – m) = 0.4n – 0.2m

Thus,
(4): log_b y = n – 3 log_b x = 0.6m – 0.2n


And hence,
log_b y/x = log_b y - log_b x = 0.8m – 0.6n
log_b √(xy) = 0.5 (log_b x + log_b y) = 0.2m + 0.1n

O lvl A Maths: Indices and Logarithms

From http://www.sgforums.com/forums/2297/topics/329124

express the following in the form ln x = ax + b and find a and b

(xe^x)² = 30e^-x

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Answer:

(xe^x)² = 30e^-x
2 ln(xe^x) = ln (30e^-x)
2 ln x +2 ln (e^x) = ln 30 + ln (e^-x)
2 ln x + 2x = ln 30 - x

so,
2 ln x = -3x + ln 30
ln x = -1.5x + 0.5 ln 30

thus,
a = -1.5
b = 0.5 ln30 = 1.7