## Slot -2 – Quantitative Aptitude – Algebra – Logarithms – If p^3 = q^4 = r^5 = s^6

If p^3 = q^4 = r^5 = s^6, then the value of log_s⁡pqr is equal to?

a) 24/5

b) 16/5

c) 47/10

d) 1

Solution:

Let p^3 = q^4 = r^5 = s^6=k

So p=k^(1/3), q=k^(1/4), r=k^(1/5) and s=k^(1/6)

Thus log(base s)⁡pqr = log(base(k^(1/6) ))⁡ k^(1/3+1/4+1/5) =6 log(base k)⁡ k^((20+15+12)/60)=6×47/60 log(base k)⁡ k=47/10

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## Slot -2 – Quantitative Aptitude – Algebra – Logarithms – 1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base5⁡)100

1/log(base2⁡)100 -1/log(base4)⁡100 +1/log(base5)⁡100 -1/log(base10)⁡100 +1/log(base20⁡)100 -1/log(base25)⁡100 +1/log(base50)⁡100 —?

a) ½

b) 10

c) -4

d) 0

Solution:
Using log(basea)⁡b = 1/log(baseb⁡)a

1/log(base2⁡)100 -1/log(base4⁡)100 +1/log(base5⁡)100 -1/log(base10)⁡100 +1/log(base20)⁡100 -1/log(base25)⁡100 +1/log(base50)⁡100

= log(base100)⁡2- log(base100)⁡4+ log(base100⁡)5-log(base100⁡)10+ log(base100⁡)20-log(base100⁡)25+log(base100⁡)50

=log(base100)⁡ (2×5×20×50)/(4×10×25)

=log(base(10^2 ))⁡10=1/2

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## Slot -1 – Quantitative Aptitude – Algebra – Logarithms – If x is a positive quantity such that 2^x

If x is a positive quantity such that 2^x = 3^log(base5)^⁡2 , then x is equal to?

a) log(base5⁡)^9

b) 1 + log(base5)⁡ 3/5

c) log(base5⁡)^8

d) 1 + log(base3) ⁡5/3

Answer: b) 1 + log(base5)⁡ 3/5

Solution:
Given , 2^x = 3^log(base5)⁡2

taking log of both sides ,

x log⁡2 = log(base5) 2 log⁡ 3 = (log⁡ 2 log⁡ 3 )/log⁡5
Or x = log 3/log⁡ 5 = log(base5)⁡ 3 = 1+ log(base5)⁡ 3/5

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## Slot -1 – Quantitative Aptitude – Algebra – Logarithms – If log(base12)⁡ 81=p then 3

If log(base12)⁡ 81=p then 3 { (4-p)/(4+p)} is equal to

a) log (base2) 8

b) log (base4) 16

c) log (base6) 8

d) log (base6) 16

Solution:

Given , log(base12)⁡ 81 = p

4 log (base12) 3 = p

So 3 (4-p)/(4+p) = 3 (1- log(base12)⁡ 3 )/(1+ log(base12)⁡ 3)=3 × log(base12)⁡(12/3)/log(base12)⁡(12×3) = 3×log(base36)⁡ 4 = log(base6) ⁡8

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## Slot -1 – Quantitative Aptitude – Algebra – Logarithms – If log(base2)(5 + log(base3) a)

If log2(5 + log3a) = 3 and log5(4a + 12 + log2 b) = 3, then a + b is equal to?

a) 59

b) 40

c) 67

d) 32

Solution:

Given, log2(5 + log3 a) = 3

5 + log3 a = 2^3 = 8

log3 a = 3

so a =  3^3 = 27

Now log(base5⁡) (4a+12+log(base2)⁡ b)=3

Or 4a + 12 + log(base2) b) = 125

log(base2)⁡ b = 125-12-4×27=5

So b = 2^5 = 32

Thus a+b = 59

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