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**Question 1: **If $\log_{32}b = \frac{d}{e}$, find the number of values of b if d,e are integers and b is a natural number less than 2000?

a) 9

b) 11

c) 10

d) 15

**Question 2:** If $\log_{16}5 = x$ and $\log_53 = y$, find the value of $\log_36$ in terms of x and y?

a) $\frac{1+4xy}{4xy}$

b) $\frac{4xy}{1+4xy}$

c) $\frac{1}{1+4xy}$

d) $1+4xy$

**Question 3:** How many values of x satisfy the equation $log_2log_3(x^4 + 4x^3 -16x + 65)$ = $6log_{125}( log_{5^{\frac{1}{2}}}125 – 1)$

a) 0

b) 1

c) 2

d) 4

**Question 4: **If x,y are natural numbers ,$log8+log5 = log(x+y)+log(x-y)$ and log8+log11 = log(x-y)+log(1+xy), then find the value of $(x^3-y^3)$

a) 325

b) 316

c) 643

d) 602

**Question 5: **If $3^{2x + 3} – 244*3^x = -9$, then which of the following statements is true?

a) ‘x’ is a positive number

b) ‘x’ is a negative number

c) ‘x’ can be either a positive number or a negative number

d) None of the above

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__Answers & Solutions:__

**1) Answer (B)**

$\log_{32}b = \frac{d}{e}$

$b = 2^{5d/e}$

Since d and e can take any integral value and b is a natural number and a power of 2

Therefore b = [$2^0,2^{10}$]. There are a total of 11 values.

**2) Answer (A)**

$\log_{16}5 = x$

$\frac{\log5}{4\log2} = x$

$\log_53 = y$

$\frac{\log3}{\log5} = y$

We have to find the value of $\log_36$ = 1+$\frac{\log2}{\log3}$ = 1+$\frac{\log5}{4xy\log5}$ = 1+1/4xy = $\frac{1+4xy}{4xy}$

**3) Answer (C)**

LHS = $6log_{125}( log_{5^{\frac{1}{2}}}125 – 1)$

= $6log_{125}(2 log_{5}125 – 1)$

= $6log_{125}(6 – 1)$

=$6log_{125}5$

=$1/3*6log_{5}5$

=2

$log_2log_3(x^4 + 4x^3 -16x + 65)$ =$ 6log_{125}(log_{5^{\frac{1}{2}}}125 – 1) = log_{125}5^6 = 2$

$log_3(x^4 + 4x^3 -16x + 65) = 4$

$x^4 + 4x^3 -16x + 65 = 81 $

$x^4 + 4x^3 -16x – 16 = 0$

$(x – 2)(x + 2)^3 = 0$

thus x can take two values 2 and -2.

**4) Answer (B)**

log(x+y)(x-y) = log40

(x+y)(x-y) = 40

x+y = 10 and x-y = 4 or x+y = 20 and x-y = 2

x = 7 and y = 3 or x = 11 and y = 9.

log88 = log(x-y)(1+xy)

x = 7 and y = 3 satisfy the equation.

$(x^3-y^3) = 7^3-3^3 = 316$

**5) Answer (C)**

The equation can be written as follow:

$3^{2x} * 3^3 – 244 * 3^x + 9 = 0$

Let $3^x = t$

=> $27t^2 – 244t + 9 = 0$

=> $27t^2 – 243t – t + 9 = 0$

=> $27t(t – 9) – 1(t – 9) = 0$

=> t = 9 or t = 1/27

=> $3^x = 9 or 3^x = 1/27$

=> $3^x = 3^2 or 3^x = 3^{-3}$

So, x = 2 or x = -3

So, ‘x’ can be either a positive number or a negative number. Option c) is the correct answer