NIMCET 2027 Logarithms And Indices Previous Year Questions (PYQs) – Page 1 of 3
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Showing 21 questions
Solve inequality
$\log_3\big((x+2)(x+4)\big)+\log_{1/3}(x+2)<\dfrac12\log_{\sqrt{3}}7$
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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 $x>-2$ ⇒ $(-2,3)$
Number of distinct integer values of $a$ satisfying
$2^{2a} - 3(2^{a+2}) + 25 = 0$ is:
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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.
is
Let:
$ X = 2^{100},\quad Y = 3^{100},\quad Z = 4^{100} $
Which statement is true?
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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 $
The value of $9^{\frac{1}{3}}.9^{\frac{1}{9}}.9^{\frac{1}{27}}..... \infty $is.
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Solution
If $log_x^y=100$ and $log_2^x=10$ then the value of y is.
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Solution
NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
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