NIMCET 2027 Logarithms And Indices Previous Year Questions (PYQs) – Page 2 of 3
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$6 + \log_{1/4} \dfrac{1}{\sqrt{2}} \left[ \sqrt{1 - \dfrac{1}{\sqrt{2}}} \sqrt{1 - \dfrac{1}{\sqrt{2}}} \sqrt{1 - \dfrac{1}{\sqrt{2}}} \cdots \right] =$
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Solution
The value of
$y=0.36\log_{0.25}\left(\dfrac13+\dfrac1{3^2}+\cdots\right)$
is
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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: $\boxed{0.18}$
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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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: $\boxed{3}$
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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Solution
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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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:
$\log_{4/5}\left( (4/5)^2 \right) = 2$
If (x0, y0) is the solution of the equations (2x)ln2 = (3y)ln3 and 3lnx = 2lny, then x0 is:
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Solution
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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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 $
If $a = log_{12}^{18}$, $b = log_{24}^{54}$, then $ab + 5(a - b)$ is
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Solution
NIMCET 2027
Online Test Series, Information About Examination,
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and More.
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NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
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