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MCA NIMCET Logarithms And Indices PYQ

Qus : 1 MCA NIMCET PYQ

Solve inequality $\log_3\big((x+2)(x+4)\big)+\log_{1/3}(x+2)<\dfrac12\log_{\sqrt{3}}7$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Domain: $x>-2$ and $x>-4$ ⇒ $x>-2$. Convert logs: $\log_{1/3}(x+2)= -\log_3(x+2)$ $\log_{\sqrt{3}}7=2\log_3 7$ RHS $=\dfrac12\cdot2\log_3 7=\log_3 7$ LHS: $\log_3((x+2)(x+4)) - \log_3(x+2)=\log_3(x+4)$ So inequality becomes $\log_3(x+4) < \log_3 7$ Thus: $x+4<7$ $x<3$ Combine with domain $x>-2$ ⇒ $(-2,3)$

Qus : 2 MCA NIMCET PYQ

Number of distinct integer values of $a$ satisfying $2^{2a} - 3(2^{a+2}) + 25 = 0$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Rewrite: $2^{2a} - 12\cdot 2^a + 25 = 0$

Let $t = 2^a$ (positive).

Equation becomes: $t^2 - 12t + 25 = 0$

$t = \dfrac{12 \pm \sqrt{44}}{2} = 6 \pm \sqrt{11}$

We need $t = 2^a$, a power of 2.

Check if $6 + \sqrt{11}$ or $6 - \sqrt{11}$ is a power of 2.

Values: $6 + \sqrt{11} \approx 9.316$ → not power of 2

$6 - \sqrt{11} \approx 2.684$ → not power of 2 No integer $a$ exists.

Qus : 3 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

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Qus : 4 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

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Qus : 5 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

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Qus : 6 MCA NIMCET PYQ

The solution set of the inequality

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 7 MCA NIMCET PYQ

Let: $ X = 2^{100},\quad Y = 3^{100},\quad Z = 4^{100} $ Which statement is true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$ Z = 4^{100} = (2^{2})^{100} = 2^{200} $ Compare: $ X = 2^{100} $ $ Y = 3^{100} $ $ X + Y = 2^{100} + 3^{100} $ Since: $ 3^{100} < 4^{100} $ Therefore: $ X + Y < Z $

Qus : 8 MCA NIMCET PYQ

The value of $9^{\frac{1}{3}}.9^{\frac{1}{9}}.9^{\frac{1}{27}}..... \infty $is.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 9 MCA NIMCET PYQ

If $log_x^y=100$ and $log_2^x=10$ then the value of y is.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 10 MCA NIMCET PYQ

The solution set of equation log_x2 log_2x 2 = log_4x 2 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 11 MCA NIMCET PYQ

The value of ${3}^{3-\log _35}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 12 MCA NIMCET PYQ

$6 + \log_{1/4} \dfrac{1}{\sqrt{2}} \left[ \sqrt{1 - \dfrac{1}{\sqrt{2}}} \sqrt{1 - \dfrac{1}{\sqrt{2}}} \sqrt{1 - \dfrac{1}{\sqrt{2}}} \cdots \right] =$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 13 MCA NIMCET PYQ

The value of $y=0.36\log_{0.25}\left(\dfrac13+\dfrac1{3^2}+\cdots\right)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

The series is a G.P. with first term $\dfrac13$ and ratio $\dfrac13$. Sum $=\dfrac{\frac13}{1-\frac13}=\dfrac12$ So $y=0.36\log_{0.25}\left(\dfrac12\right)$ $\log_{0.25}\left(\dfrac12\right)=\dfrac{\log(1/2)}{\log(1/4)}=\dfrac{-1}{-2}=\dfrac12$ Hence $y=0.36\times\dfrac12=0.18$ Answer: $\boxed{0.18}$

Qus : 14 MCA NIMCET PYQ

For $a>0,\ a\ne1$, the number of values of $x$ satisfying $2\log_x a+\log_{ax} a+3\log_{a^2x} a=0$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Let $\log_x a=t$ Then $\log_{ax}a=\dfrac{t}{1+t}$, $\log_{a^2x}a=\dfrac{t}{2+t}$ Equation becomes $2t+\dfrac{t}{1+t}+3\dfrac{t}{2+t}=0$ Solving gives $t=0,-1,-2$ All give valid $x$ values. Number of solutions $=3$ Answer: $\boxed{3}$

Qus : 15 MCA NIMCET PYQ

If $\log (1-x+x^2)={{a}}_1x+{{a}}_{2{}^{{}^{}}}{x}^2+{{{}{{a}}_3{x}^3+.\ldots.}}^{}$ then ${{a}}_3+{{a}}_6+{{a}}_9+.\ldots.$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 16 MCA NIMCET PYQ

If $8^{x-1}=(1/4)^{x}$, then the value of $\frac{1}{\log_{x+1}4-\log_{x+1}5}+\frac{1}{\log_{1-x}4-\log_{1-x}5}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given: $8^{x-1} = (1/4)^{x}$

Rewrite both sides with base 2: $8 = 2^3$ and $1/4 = 2^{-2}$

So: $(2^3)^{x-1} = (2^{-2})^x$ $\Rightarrow 2^{3(x-1)} = 2^{-2x}$

Equate powers: $3(x - 1) = -2x$

$3x - 3 = -2x$

$5x = 3$

$\Rightarrow x = \frac{3}{5}$

We need the value of:

$\displaystyle \frac{1}{\log_{x+1} 4 - \log_{x+1} 5} + \frac{1}{\log_{1-x} 4 - \log_{1-x} 5}$

Use property: $\log_a m - \log_a n = \log_a \left(\frac{m}{n}\right)$

So the expression becomes: $\displaystyle \frac{1}{\log_{x+1} \left(\frac{4}{5}\right)} + \frac{1}{\log_{1-x} \left(\frac{4}{5}\right)}$

Now use: $\displaystyle \frac{1}{\log_a b} = \log_b a$

So expression becomes: $\log_{4/5}(x+1) + \log_{4/5}(1-x)$

Use product property: $\log_{4/5}[(x+1)(1-x)]$

Compute: $(x+1)(1-x) = 1 - x^2$

Substitute $x = \frac{3}{5}$:

$1 - x^2 = 1 - \frac{9}{25} = \frac{16}{25}$

Thus value = $\log_{4/5}\left(\frac{16}{25}\right)$

Rewrite: $\frac{16}{25} = \left(\frac{4}{5}\right)^2$

Therefore: $\log_{4/5}\left( (4/5)^2 \right) = 2$

Qus : 17 MCA NIMCET PYQ

If (x₀, y₀) is the solution of the equations (2x)^ln2 = (3y)^ln3 and 3^lnx = 2^lny, then x₀ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 18 MCA NIMCET PYQ

Let $x$ be a positive real number such that $x^{(8\log_5x-24)}=5^{-4}$. Then the product of all possible values of x is =

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

We have $ x^{8\log_5 x - 24} = 5^{-4} $

Let $ \log_5 x = t $.

Then $ x = 5^t $.

Substitute: $ (5^t)^{8t - 24} = 5^{-4} $

$ 5^{t(8t - 24)} = 5^{-4} $

So: $ t(8t - 24) = -4 $

$ 8t^2 - 24t + 4 = 0 $

Divide by 4:

$ 2t^2 - 6t + 1 = 0 $

Solutions: $ t = \frac{3 \pm \sqrt{7}}{2} $

Thus: $ x_1 = 5^{\frac{3 + \sqrt{7}}{2}},\ x_2 = 5^{\frac{3 - \sqrt{7}}{2}} $

Product: $ x_1 x_2 = 5^{\frac{3 + \sqrt{7}}{2} + \frac{3 - \sqrt{7}}{2}} = 5^{\frac{6}{2}} = 5^3 = 125 $

Qus : 19 MCA NIMCET PYQ

If $3^x = 4^{x-1}$, then $x=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

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Qus : 20 MCA NIMCET PYQ

If $a = log_{12}^{18}$, $b = log_{24}^{54}$, then $ab + 5(a - b)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

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Qus : 21 MCA NIMCET PYQ

If $x, y, z$ are three consecutive positive integers, then $log (1 + xz)$ is

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