Integer Answer Type Questions for JEE: Logarithms

Q.1. The number when simplified reduces to a natural number N. Find N.

Ans. 2

= (3 + log₅2) (2 +log₅2) - (1 + log₅2) (4 + log₅2)
= (log₅2)² + 5 log₅2 + 6 - [(log₅2)² + 5 log₅2 + 4]= 2

Q.2. The number N = 6log₁₀2 + log₁₀31 lies between two successive integers whose sum equals_______

Ans. 7
∴ N = log₁₀64 + log₁₀31 = log₁₀1984
∴ 3 < N < 4

Q.3. Suppose x, y, z > 0 & different from one and λn  x + λn y + λn z = 0, then value of
is e ^{- k} then k equals _____

Ans. 3
Let X
⇒ λn x = λn (λn z)
Now given λn x + λn y + λn z = 0

Similarly,
 & 

∴ R.H.S. = - 3
∴ λn X = - 3
X =  e^-3

Q.4. If log (x²/y³) = 1 & log(x²y³) = 7 then log |xy| is equal to _____

Ans. 3
2logx - 3log y = 1
2log x + 3log y = 7
Solving  log x  = 2
& log y  = 1   ∴ log x + log y = log xy = 3

Q.5. The number of positive integers satisfying x + log₁₀ (2^x + 1) = x log₁₀5 + log₁₀6 is .......

Ans. 1
x (log₁₀10 - log₁₀5) + log₁₀(2^x + 1) = log₁₀6
⇒ x log₁₀2 + log₁₀ (2^x + 1) = log₁₀6
⇒ log₁₀2^x (2^x + 1) = log₁₀6
⇒ 2^2x + 2^x - 6 = 0
⇒ (2^x + 3) (2^x - 2) = 0
⇒ ∴ 2x + 3 ≠ 0          ∴ 2^x = 2
⇒  x = 1

Q.6. If x, y, z be positive real numbers such that log_2x z = 3, log_5y z = 6 and log_xy z = 2/3 then the value of z is in the form of m/n in lowest form then find value of n - m.

Ans. 9
z = 8x³,  z = 5⁶y⁶, z = x^2/3y^2/3

Q.7. Let P = log₅ (log₅ 3). If ₃C+5^-P = 405 then C equals

Ans. 4
∴ 5- P = ₅- log₅ log₅ (3) ⇒ 5^-P = log3⁵ ⇒ ₃C + log₃ 5 = 405
(3^C) (5) = 405 ⇒  3^C = 81 ⇒ C = 4

Q.8. Number of solution for |3x² - 2 | = [-2π] is ([.] denotes greatest integer)

Ans. 0
[- 2π] = - 7        ∴ L.H.S. is positive
R.H.S. is negative     ∴ no solution

Q.9. Find all positive values of 'a' for which the equation log (ax) = 2 log (x + 1) has the unique root.

Ans. 4
log ax = 2 log (x + 1)
⇒ ax = (x + 1)² ⇒ x² + (2 -a) x + 1 = 0
Let x₁ and x₂ be roots
and

For solution exists a² -4a ≥ 0
a ≥ 4, a ≤ 0
For positive values a ≥ 4
For a > 4, x₁ and x₂ are different
For a = 4, x₁ and x₂ are same
Hence a = 4.

Q.10. If no. of zeroes after decimal in (0.15)²⁰ is ab. Find b - a.
(Assume log 2 = 0.3010, log 3 = 0.4771)   

Ans. 5
log x = 20 (log 0.15)
= 20 [1 - log 2 + log 3 - 2]
= 20 [-1 + 0.176] = - 20 + 352 = - 17 + 0.52
c = - 17      no. of zeroes = | - 17 + 1 | = 16
ab = 16    ∴ b - a = 5