Logarithms

Let x be the required number.
Then, log 0.0001 x = -1/4
-> x = (0.0001)^(-1/4) = ( 1/10000)^(-1/4) = 10000^(1/4) = 10

Let log ₇ (343√7) = x
-> 7^x = 343√7 = 7³ (7^½)
-> 7^x = 7^7/2
or x = 7/2 = 3.5

Log₂(log₂x) = 4
-> Log₂x = 2⁴ = 16
-> x = 2¹⁶

Alternatively
You may also proceed by plugging in values of x in different options, and checking for the right fit.
log₂(log₂2¹⁶) = log₂(16log₂ 2) = (log₂ 16) = log₂ 2⁴ = 4

Let log₂x = k
-> (k-2)/(k-4) < 0
-> 2 < k < 4
-> 2 < log₂x < 4
-> 4 < x < 16
Maximum m-n = 10, when m=15, n= 5
Maximum m+n = 29, when one is 15 and the other is 14.

 

Given, log₅ (5^1/k+5³) = log₅ 6+1+1/(2k)
-> log₅ (5^1/k+5³) = log₅ 6 + log₅ 5 + log₅ 5^1/2k
-> Log₅(5^1/k + 5³) = log₅(30*5^1/2k)
-> 5^1/k + 125 = 30*5^1/2k
Put 5^1/2k = m à m²+125 = 30 m
-> m²-30m+125 = 0
-> (m-5)(m-25) = 0
-> m = 5, 25
-> 5^1/2k = 5¹ or 5²
-> 1/2k = 1 or 2
-> k = ½ or ¼

Given, 2logx+logx3 = log32
-> logx²+3logx = log4 + log8
-> logx²+logx³ = log2²+log2³
-> x = 2 (comparing L.H.S and R.H.S)

Alternatively
Put values to verify that 2 fits the equation.

If a, b, c are in A.P., then b-a=c-b or 2b = a+c
Thus, 2 log₃(2^x-5) = log₃2 + log₃(2^x-7/2)
-> log₃(2^x-5)² = log₃ 2(2^x-7/2)
-> log₃(2^x-5)² = log₃ (2^x+1-7)
-> (2^x-5)² = 2^x+1-7
Put 2^x = y
-> (y-5)² = 2y-7
-> (y-4)(y-8) = 0
-> y = 4, 8
If y = 4, then 2^x = 4 à x = 2
If y = 8, then 2^x = 8 à x = 3

Alternatively
Put values to verify that 3 fits the equation.

 

 

2^{(2 – log} ₂ ⁷⁾ = 2^((2log ₂ ^{2) – log} ₂ ⁷⁾   ∵ log ₂ 2 = 1
= 2^(log ₂ ^{4 – log} ₂ ⁷⁾
= 2^log ₂^{(4/ 7)}
= 4/7     ∵ a^(log _a ⁿ⁾ = n