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# Logarithms & Indices Questions For CAT Set-3 PDF:

Download Logarithms & Indices Questions For CAT Set-3 PDF. Practice important problems on Logarithms & Indices Questions  with detailed explanations and solutions.

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

$\log_{32}b = \frac{d}{e}$
$b = 2^{5d/e}$
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.

$\log_{16}5 = x$
$\frac{\log5}{4\log2} = x$
$\log_53 = y$
$\frac{\log3}{\log5} = y$
We have to find the value of $\log_36$ = 1+$\frac{\log2}{\log3}$ = 1+$\frac{\log5}{4xy\log5}$ = 1+1/4xy = $\frac{1+4xy}{4xy}$

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.

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

$3^{2x} * 3^3 – 244 * 3^x + 9 = 0$
Let $3^x = t$
=> $27t^2 – 244t + 9 = 0$
=> $27t^2 – 243t – t + 9 = 0$
=> $27t(t – 9) – 1(t – 9) = 0$
=> $3^x = 9 or 3^x = 1/27$
=> $3^x = 3^2 or 3^x = 3^{-3}$