# Discrete Mathematics Questions and Answers – Discrete Probability – Logarithmic Series

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This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Discrete Probability – Logarithmic Series”.

1. Computation of the discrete logarithm is the basis of the cryptographic system _______
a) Symmetric cryptography
b) Asymmetric cryptography
c) Diffie-Hellman key exchange
d) Secret key cryptography

Explanation: A discrete logarithm modulo of an integer to the base is an integer such that ax ≡ b (mod g). The problem of computing the discrete logarithm is a well-known challenge in the field of cryptography and is the basis of the cryptographic system i.e., the Diffie-Hellman key exchange.

2. Solve the logarithmic function of ln($$\frac{1+5x}{1+3x}$$).
a) 2x – 8x2 + $$\frac{152x^3}{3}$$ – …
b) x2 + $$\frac{7x^2}{2} – \frac{12x^3}{5}$$ + …
c) x – $$\frac{15x^2}{2} + \frac{163x^3}{4}$$ – …
d) 1 – $$\frac{x^2}{2} + \frac{x^4}{4}$$ – …

Explanation: To solve the logarithmic function ln($$\frac{1+5x}{1+3x}$$) = ln(1+5x) – ln(1+3x) = (5x – $$\frac{(5x)^2}{2} + \frac{(5x)^3}{3}$$ – …) – (3x – $$\frac{(3x)^2}{2} + \frac{(3x)^3}{3}$$ – …) = 2x – 8x2 + $$\frac{152x^3}{3}$$ – …

3. Determine the logarithmic function of ln(1+5x)-5.
a) 5x + $$\frac{25x^2}{2} + \frac{125x^3}{3} + \frac{625x^4}{4}$$ …
b) x – $$\frac{25x^2}{2} + \frac{625x^3}{3} – \frac{3125x^4}{4}$$ …
c) $$\frac{125x^2}{3} – 625x^3 + \frac{3125x^4}{5}$$ …
d) -25x + $$\frac{125x^2}{2} – \frac{625x^3}{3} + \frac{3125x^4}{4}$$ …

Explanation: Apply the logarithmic law, that is logax = xlog(a). Now the function is ln(1+5x)-5 = -5log(1+5x). By taking the series = -5(5x – $$\frac{(5x)^2}{2} + \frac{(5x)^3}{3} – \frac{(5x)^4}{4}$$ + …) = -25x + $$\frac{125x^2}{2} – \frac{625x^3}{3} + \frac{3125x^4}{4}$$ …

4. Find the value of x: 3 x2 alogax = 348?
a) 7.1
b) 4.5
c) 6.2
d) 4.8

Explanation: Since, alogax = x. The given equation may be written as: 3x2 x = 348 ⇒ x = (116)1/3 = 4.8.

5. Solve for x: log2(x2-3x)=log2(5x-15).
a) 2, 5
b) 7
c) 23
d) 3, 5

Explanation: By using the property if logax = logay then x=y, gives 2x2-3x=10-6x. Now, to solve the equation x2-3x-5x+15=0 ⇒ x2-8x+15 ⇒ x=3, x=5
For x=3: log2(32-3*3) = log2(5*3-15) ⇒ true
For x=5: log2(52-3*5) = log2(5*5-15) ⇒ true
The solutions to the equation are : x=3 and x=5.

6. Solve for x the equation 2x + 3 = 5x + 2.
a) ln (24/8)
b) ln (25/8) / ln (2/5)
c) ln (32/5) / ln (2/3)
d) ln (3/25)

Explanation: Given that 2x + 3 = 5x + 2. By taking ln of both sides: ln (2x + 3) = ln (5x + 2)
⇒ (x + 3) ln 2 = (x + 2) ln 5
⇒ x ln 2 + 3 ln 2 = x ln 5 + 2 ln 5
⇒ x ln 2 – x ln 5 = 2 ln 5 – 3 ln 2
⇒ x = ( 2 ln 5 + 3 ln 2 ) / (ln 2 – ln 5) = ln (52 / 23) / ln (2/5) = ln (25/8) / ln (2/5).

7. Given: log4 z = B log2/3z, for all z > 0. Find the value of constant B.
a) 2/(3!*ln(2))
b) 1/ln(7)
c) (4*ln(9))
d) 1/(2*ln(3))

Explanation: By using change of base formula we can have ln (x) / ln(4) = B ln(x) / ln(2/3) ⇒
B = 1/(2*ln(3)).

8. Evaluate: 16x – 4x – 9 = 0.
a) ln [( 5 + $$\sqrt{21}$$) / 2] / ln 8
b) ln [( 2 + $$\sqrt{33}$$) / 2] / ln 5
c) ln [( 1 + $$\sqrt{37}$$) / 2] / ln 4
d) ln [( 1 – $$\sqrt{37}$$) / 2] / ln 3

Explanation: Given: 16x – 4x – 9 = 0. Since 16x = (4x)2, the equation may be written as: (4x)2 – 4x – 9 = 0. Let t = 3x and so t: t2 – t – 9 = 0 which gives t: t = (1 + $$\sqrt{37}$$) / 2 and (1 – $$\sqrt{37}$$) / 2
Since t = 4x, the acceptable solution is y = (1 + $$\sqrt{37}$$) / 2 ⇒ 4x = (1 + $$\sqrt{37}$$)/2. By using ln on both sides: ln 4x = ln [ (1 + $$\sqrt{37}$$) / 2] ⇒ x = ln [ ( 1 + $$\sqrt{37}$$)/2] / ln 3.

9. Transform 54y = n+1 into equivalent a logarithmic expression.
a) log12 (n+1)
b) log41 (n2)
c) log63 (n)
d) log54 (n+1)

Explanation: By using the equivalent expression: ay = x ⇔ y = loga (x) to write 3x = m as a logarithm: y = log54 (n+1).

10. If loga$$(\frac{1}{8}) = -\frac{3}{4}$$, than what is x?
a) 287
b) 469
c) 512
d) 623