JEE MAIN Logarithms And Indices Previous Year Questions (PYQs) – Page 1 of 2
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The sum of all the real solutions of the equation
$\log_{x+3}(6x^2 + 28x + 30) - 5 = 2\log_{6x+10}(x^2 + 6x + 9)$
is equal to:
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Solution
$\log_{x+3}((x+3)(6x+10)) = 5 - 2\log_{6x+10}(x+3)^2$
$1 + \log_{x+3}(6x+10) = 5 - 4\log_{6x+10}(x+3)$
Let $\log_{x+3}(6x+10) = A$
$\Rightarrow A + \frac{4}{A} = 4 \Rightarrow A = 2$
$\Rightarrow \log_{x+3}(6x+10) = 2$
$6x + 10 = (x+3)^2$
$x^2 - 1 = 0 \Rightarrow x = \pm 1$
Sum of roots $= 0$
If $\log_e a,\ \log_e b,\ \log_e c$ are in an A.P. and
$\log_e a-\log_e 2b,\ \log_e 2b-\log_e 3c,\ \log_e 3c-\log_e a$ are also in an A.P.,
then $a:b:c$ is equal to:
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Solution
If for x $\in$ $\left( {0,{\pi \over 2}} \right)$, log10sinx + log10cosx = $-$1 and log10(sinx + cosx) = ${1 \over 2}$(log10 n $-$ 1), n > 0, then the value of n is equal to :
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Solution
For three positive integers $p, q, r$, $x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ and $r = pq + 1$ such that
$3,\ 3\log_{y}x,\ 3\log_{z}y,\ 7\log_{x}z$ are in A.P. with common difference $\dfrac{1}{2}$.
Then $r - p - q$ is equal to:
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Solution
The number of integral solutions $x$ of $\log _{\left(x+\frac{7}{2}\right)}\left(\frac{x-7}{2 x-3}\right)^{2} \geq 0$ is :
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Solution
If the domain of the function
$f(x)=\log_{(10x^2-17x+7)}(18x^2-11x+1)$
is $(-\infty,a)\cup(b,c)\cup(d,\infty)-{e}$, then $90(a+b+c+d+e)$ equals:
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Solution
For logarithm:
$18x^2 - 11x + 1 > 0$
$(2x-1)(9x-1)>0$
$x<\frac{1}{9}$ or $x>\frac{1}{2}$
Also base condition:
$10x^2 - 17x + 7 > 0$
$(x-1)(10x-7)>0$
$x<\frac{7}{10}$ or $x>1$
Also
$10x^2 -17x +7 \ne 1$
$\Rightarrow x \ne \frac{6}{5}$
Combine intervals
$x\in(-\infty,\frac{1}{9})\cup(\frac{1}{2},\frac{7}{10})\cup(1,\infty)\setminus\left{\frac{6}{5}\right}$
So
$a=\frac{1}{9},; b=\frac{1}{2},; c=\frac{7}{10},; d=1,; e=\frac{6}{5}$
$90(a+b+c+d+e)$
$=90\left(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{6}{5}\right)$
$=10+45+63+90+108=316$
The sum of all the solutions of the equation $(8)^{2x}-16\cdot(8)^x+48=0$ is:
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Solution
Let $y=\log_e!\left(\dfrac{1-x^2}{1+x^2}\right)$, with $-1
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Solution
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