Basic Mathematics for JEE Main – Logarithmic Inequality

Q 1 :

The number of solutions of $\log_{4} (x - 1) = \log_{2} (x - 3)$ is

3
1
2
0

1

2

3

4

(2)

$For the given equation to be valid, we must have$ $x - 1 > 0 and x - 3 > 0$

$\Rightarrow$ $x > 3$

$We can write the given equation as \frac{\log (x - 1)}{\log 4} = \frac{\log (x - 3)}{\log 2}$

$\Rightarrow$ $\log (x - 1) = 2 \log (x - 3)$ $(∵ \log 4 = 2 \log 2)$

$\Rightarrow$ $(x - 1) = {(x - 3)}^{2} \Rightarrow x^{2} - 7 x + 10 = 0$

$\Rightarrow$ $(x - 5) (x - 2) = 0 \Rightarrow x = 5 or 2$

$As x > 3,$ $x = 5$

Q 2 :

The value of $\log_{2} \log_{2} \log_{4} 256 + 2 \log_{\sqrt{2}} 2$ is

5
2
3
4

1

2

3

4

(1)

$\log_{2} \log_{2} \log_{4} 256 + 2 \log_{\sqrt{2}} 2$

$= \log_{2} \log_{2} \log_{4} 4^{4} + \frac{2 \log 2}{\log \sqrt{2}}$ $[∵ \log_{x} x = 1]$

$= \log_{2} \log_{2} 4 + \frac{2 \log 2}{\frac{1}{2} \log 2}$

$= \log_{2} 2 + 4 = 1 + 4 = 5$

Q 3 :

Let $n = 2006!$ then $\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \dots + \frac{1}{\log_{2006} n} =$

2006
2005
2005!
1

1

2

3

4

(4)

$\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \dots + \frac{1}{\log_{2006} n}$

$= \frac{\log 2}{\log n} + \frac{\log 3}{\log n} + \dots + \frac{\log (2006)}{\log n}$

$= \frac{1}{\log n} [\log 2 + \log 3 + \dots + \log 2006]$

$= \frac{1}{\log n} [\log (1 \cdot 2 \cdot 3 \dots 2006)]$

$= \frac{1}{\log n} \log (2006!) = \frac{1}{\log n} \times \log n = 1$ $(∵ n = 2006!)$

Q 4 :

If $\log_{\sqrt{3}} 5 = a$ and $\log_{\sqrt{3}} 2 = b$ , then $\log_{\sqrt{3}} 300$

$2 (a + b)$
$2 (a + b + 1)$
$2 (a + b + 2)$
$a + b + 4$

1

2

3

4

(2)

$\log_{\sqrt{3}} 5 = a,$ $\log_{\sqrt{3}} 2 = b$

$\log_{\sqrt{3}} 5 + \log_{\sqrt{3}} 2 = a + b,$ $\log_{\sqrt{3}} 10 = a + b$

$\log_{\sqrt{3}} 300 = \log_{\sqrt{3}} (3 \cdot 100) = \log_{\sqrt{3}} 3 + \log_{\sqrt{3}} {(10)}^{2}$

$= \log_{\sqrt{3}} {(\sqrt{3})}^{2} + 2 \log_{\sqrt{3}} 10 = 2 + 2 (a + b) = 2 (a + b + 1)$

Q 5 :

The value of $\log_{2} 20 \log_{2} 80 - \log_{2} 5 \log_{2} 320$ is equal to

5
6
7
8

1

2

3

4

(4)

$\log_{2} 20 \log_{2} 80 - \log_{2} 5 \log_{2} 320$

$\log_{2} 20 \log_{2} 80 - \log_{2} 5 \log_{2} (20 \times 16)$

$= \log_{2} 20 \log_{2} (16 \times 5) - \log_{2} 5 \log_{2} (20 \times 16)$

$= \log_{2} 20 (\log_{2} 16 + \log_{2} 5) - \log_{2} 5 (\log_{2} 20 + \log_{2} 16)$

$= \log_{2} 20 \log_{2} 16 + \log_{2} 20 \log_{2} 5 - \log_{2} 5 \log_{2} 20 - \log_{2} 5 \log_{2} 16$

$= \log_{2} 16 (\log_{2} 20 - \log_{2} 5) + 0$

$= \log_{2} 16 \log_{2} \frac{20}{5} = \log_{2} 16 \log_{2} 4$

$= \log_{2} 2^{4} \log_{2} 2^{2} = 4 \times 2 = 8$ $(∵ \log_{2} 2 = 1)$

Q 6 :

$\log_{4 \sqrt[3]{4^{2}}} (\frac{1}{1024}) =$

- 3
3
- 5
5

1

2

3

4

(1)

$We have,$ $\log_{4 \sqrt[3]{4^{2}}} (\frac{1}{1024})$

$= \frac{\log (\frac{1}{1024})}{\log (4 (4^{2})^{1 / 3})} = \frac{- \log {(2)}^{10}}{\log 4^{5 / 3}}$

$= \frac{- \log {(2)}^{10}}{\log {[(2)^{2}]}^{5 / 3}} = \frac{- 10 \log 2}{\frac{10}{3} \log 2} = - 3$

Q 7 :

The value of $6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots}}})$

4
3
2
1

1

2

3

4

(1)

$Let$ $t = \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots}}$

$t^{2} = 4 - \frac{1}{3 \sqrt{2}} t \Rightarrow t^{2} + \frac{1}{3 \sqrt{2}} t - 4 = 0 \Rightarrow t = \frac{8}{3 \sqrt{2}}$

$The required answer$

$= 6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \times \frac{8}{3 \sqrt{2}}) = 6 + \log_{\frac{3}{2}} \frac{4}{9} = 6 - 2 = 4$

Q 8 :

The value of $({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$ is ______ .

(8)

$({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$

$= {(\log_{2} 9)}^{\frac{2}{\log_{2} (\log_{2} 9)}} \cdot {(\sqrt{7})}^{\log_{7} 4}$

$= {(\log_{2} 9)}^{2 \log_{\log_{2} 9} (2)} \cdot 7^{\frac{1}{2} \log_{7} 4}$

$= 4 \cdot 2 = 8$

Q 9 :

The product of all positive real values of $x$ satisfying the equation $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$ is _______ .

(1)

We have, $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$

Taking log both sides, we get

$\Rightarrow$ $t (16 t^{3} - 68 t) = - 16$ $[\log_{5} x = t]$

$\Rightarrow$ $16 t^{4} - 68 t^{2} + 16 = 0 \Rightarrow 4 t^{4} - 17 t^{2} + 4 = 0$

$∵$ $t_{1} + t_{2} + t_{3} + t_{4} = 0$

$\Rightarrow$ $\log_{5} x_{1} + \log_{5} x_{2} + \log_{5} x_{3} + \log_{5} x_{4} = 0 \Rightarrow x_{1} x_{2} x_{3} x_{4} = 1$

Q 10 :

If $\log_{12} 27 = a$ , then $\log_{6} 16 =$

$2 (\frac{3 - a}{3 + a})$
$3 (\frac{3 - a}{3 + a})$
$4 (\frac{3 - a}{3 + a})$
$4 (\frac{2 - a}{2 + a})$

1

2

3

4

(3)

$a = \frac{3}{1 + 2 \log_{3} 2} \Rightarrow \log_{3} 2 = \frac{3 - a}{2 a};$ $\log_{6} 16 = \frac{4 \log_{3} 2}{1 + \log_{3} 2}$

Topic Question Set

Q 1 :

The number of solutions of $\log_{4} (x - 1) = \log_{2} (x - 3)$ is

Q 2 :

The value of $\log_{2} \log_{2} \log_{4} 256 + 2 \log_{\sqrt{2}} 2$ is

Q 3 :

Let $n = 2006!$ then $\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \dots + \frac{1}{\log_{2006} n} =$

Q 4 :

If $\log_{\sqrt{3}} 5 = a$ and $\log_{\sqrt{3}} 2 = b$ , then $\log_{\sqrt{3}} 300$

Q 5 :

The value of $\log_{2} 20 \log_{2} 80 - \log_{2} 5 \log_{2} 320$ is equal to

Q 6 :

$\log_{4 \sqrt[3]{4^{2}}} (\frac{1}{1024}) =$

Q 7 :

The value of $6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots}}})$

Q 8 :

The value of $({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$ is ______ .

Q 9 :

The product of all positive real values of $x$ satisfying the equation $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$ is _______ .

Q 10 :

If $\log_{12} 27 = a$ , then $\log_{6} 16 =$

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Topic Question Set

Q 1 : The number of solutions of log4(x-1)=log2(x-3) is

Q 2 : The value of log2log2log4256+2log22 is

Q 3 : Let n=2006! then 1log2n+1log3n+⋯+1log2006n=

Q 4 : If log35=a and log32=b, then log3300

Q 5 : The value of log220 log280-log25 log2320 is equal to

Q 6 : log4423(11024)=

Q 7 : The value of 6+log32(1324-1324-1324-132⋯)

Q 8 : The value of ((log29)2)1log2(log29)×(7)1log47 is ______ .

Q 9 : The product of all positive real values of x satisfying the equation x(16(log5x)3-68log5x)=5-16 is _______ .

Q 10 : If log1227=a, then log616=

Want Us to Email you About Special Offers & Updates?

Q 1 :

The number of solutions of $\log_{4} (x - 1) = \log_{2} (x - 3)$ is

Q 2 :

The value of $\log_{2} \log_{2} \log_{4} 256 + 2 \log_{\sqrt{2}} 2$ is

Q 3 :

Let $n = 2006!$ then $\frac{1}{\log_{2} n} + \frac{1}{\log_{3} n} + \dots + \frac{1}{\log_{2006} n} =$

Q 4 :

If $\log_{\sqrt{3}} 5 = a$ and $\log_{\sqrt{3}} 2 = b$ , then $\log_{\sqrt{3}} 300$

Q 5 :

The value of $\log_{2} 20 \log_{2} 80 - \log_{2} 5 \log_{2} 320$ is equal to

Q 6 :

$\log_{4 \sqrt[3]{4^{2}}} (\frac{1}{1024}) =$

Q 7 :

The value of $6 + \log_{\frac{3}{2}} (\frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \sqrt{4 - \frac{1}{3 \sqrt{2}} \dots}}})$

Q 8 :

The value of $({{(\log_{2} 9)}^{2})}^{\frac{1}{\log_{2} (\log_{2} 9)}} \times {(\sqrt{7})}^{\frac{1}{\log_{4} 7}}$ is ______ .

Q 9 :

The product of all positive real values of $x$ satisfying the equation $x^{(16 {(\log_{5} x)}^{3} - 68 \log_{5} x)} = 5^{- 16}$ is _______ .

Q 10 :

If $\log_{12} 27 = a$ , then $\log_{6} 16 =$