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Explore popular questions from Logarithms for SSC. This collection covers Logarithms previous year SSC questions hand picked by experienced teachers.

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Q 1. If {tex} log3 \times log4\ x > 0 {/tex}, then

A

{tex} x > 1 {/tex}

{tex} x > 4 {/tex}

C

{tex} x > 64 {/tex}

D

none of these

Explanation

We have log3 log4 x > 0

=> {tex} log4x > 1=> x > 4 {/tex}

=> x > 4

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Q 2. {tex} log \frac{1}{4} \left(a^{2}-1\right) < log_\frac{1}{2} \left(a+1\right)^{2}{/tex}

A

{tex} a<1 {/tex}

{tex} 2a + 2 < 0 => a < -1 {/tex}

C

{tex} a>1 {/tex}

D

none of these

Explanation

{tex} log_{\frac{1}{4}} \left(a^{2}-1\right) < log_{ \left(\frac{1}{4}\right) } \left(a+1\right)^{2} {/tex}

=> {tex} a^{2}-1 > \left(a+1\right)^{2} {/tex}

=> {tex} a^{2}-1 > a^{2}+1+2a {/tex}

=> {tex} 2a + 2 < 0 => a < -1 {/tex}

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Q 3. If {tex} log_{e} \left(\frac{a+b}{2}\right) = \frac{1}{2} \left(log_{e} a+log_{e} b\right) {/tex} then:

a = b

B

a = b/2

C

2a = b

D

{tex} a = \frac{b}{3} {/tex}

Explanation

{tex} log \left(\frac{a+b}{2}\right) = \frac{1}{2} \left(log_{e}a + log_{e}b\right) {/tex}

=> {tex} \frac{a+b}{2} = \sqrt{ab} {/tex}

=> {tex} a+b-2\sqrt{ab} = 0 {/tex}

=> {tex} \sqrt{a} = \sqrt{b} => a = b {/tex}

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Q 4. If {tex} 2 log \left(x+1\right)-10g \left(x^{2}-1\right) = log^{2} {/tex}, then x equals

A

1

B

0

C

2

3

Explanation

By definition x ≠ 1, =1

Given equation can be written as

{tex} log \frac{ \left(x+1\right)^{2} }{x^{2}-1} = log2 => \frac{x+1}{x-1}=2 => x = 3{/tex}

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Q 5. The number {tex} log 2^{7} {/tex} is:

A

an integer

B

a rational number

an irrational number

D

a prime number

Explanation

Let {tex} x = log 2^{7} {/tex}

=> {tex} 2^{x} = 7 {/tex}

which is only possible for irrational number

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Q 6. If {tex} y = 2^{\frac{1}{logx}\left(8\right)} {/tex}, then x is equal to:

A

y

B

{tex} y^{2} {/tex}

{tex} y^{3} {/tex}

D

none of these

Explanation

{tex} y = 2^{\frac{1}{log}x \left(8\right) } {/tex}

=> {tex} y = 2 ^{log 8x} => y = 2^{log 2\sqrt[3]{x}} {/tex}

=> {tex} x = y^{3} {/tex}

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Q 7. If {tex} log_{0.5} \sin x = 1 - log_{0.5} \cos x {/tex}, then number of solution of {tex} x ? \left[ -2 \pi , 2 \pi \right] {/tex} is:

A

1

2

C

3

D

4

Explanation

{tex} log\ _{0.5} \sin x + log\ _{0.5} \cos x = 1 {/tex}

=> {tex} log\ _{0.5} \sin x \cos x = 1 {/tex}

=> {tex} \sin x \cos x = \frac{1}{2} => \sin 2x = 1 {/tex}

=> {tex} 2x = n \pi + \left(-1\right)^{n}\frac{ \pi }{2} => x = \frac{n \pi }{2}+ \left(-1\right)^{n}\frac{ \pi }{4} {/tex}

Since {tex} log_{0.5} \sin x\ and\ log_{0.5}.\cos x\ are\ real {/tex}

{tex} \therefore \ \sin x \ and \ \cos x \ must \ lie\ in\ first\ quardrant {/tex}

{tex} \therefore x = \frac{ \pi }{4}, -2 \pi + \frac{ \pi }{4} {/tex}

Hence, the number of solutions is 2.

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Q 8. If a, b, c are in GP. then {tex} log _{ax}{x}, log_{bx}x, log_{cx}x {/tex} are in:

A

GP

HP

C

AP

D

none of these

Explanation

{tex} log_{ax}x = \frac{1}{log_{x}ax} => log_{ax}x = \frac{1}{1+log_{x}a}{/tex}

If a,b,c are in GP

=> {tex} log_{x}a, log_{x}b, log_{x}c\ are\ in\ AP {/tex}

=> {tex} 1+log_{x}a,1+log_{x}b,1+log_{x}c\ are\ in\ AP {/tex}

=> {tex} \frac{1}{1+log_{x}a}, \frac{1}{1+log_{x}b}, \frac{1}{1+log_{x}c}\ are\ in\ HP {/tex}

=> {tex} log_{ax}x,\ log_{bx}x\ log_{cx}x\ are\ in\ HP. {/tex}

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Q 9. Consider the following statements:
1. Solution of the inequality {tex} log_{5} \left(x^{2}-11x+43\right)<2 {/tex} is (0, 2)
2. If {tex} [x - 1]^{ \left(log_{3}x^{2} - 2Logx^{9}\right)} = \left(x - 1\right)^{7} {/tex}, then {tex}x{/tex} is 2 and 81.
Which of these is/are correct?

A

only (1)

only 2

C

both of these

D

none of these

Explanation

{tex} alog_{5} \left(x^{2}-11x+43\right)<2 {/tex} and {tex} x^{2}-11x+43>0 {/tex}
=> {tex} x^{2}-11x+43 < 5^{2}{/tex} and {tex} \left(x - \frac{11}{2}\right)^{2}+\frac{51}{4}>0 {/tex}
=> {tex} x^{2}-11x+18<0 {/tex} and {tex} \left(x-2\right) \left(x-9\right)<0 {/tex}
Solution is {tex} \left(2, 9\right) {/tex}
For domain {tex} |x, -1| ≠ 0, ≠ -1 {/tex}
Now {tex} |x-1| = 1, => x - 1 = \pm 1 {/tex}
=> {tex}x{/tex} = 0, 2
{tex}x{/tex} = 0 is not in the domain and {tex}x{/tex} = 2 satisfies the given equation.
If {tex}x{/tex}-1=> 0 i.e., {tex}x{/tex} => 1 then the given equation becomes
{tex} 2 log_{3}x - \frac{4}{log_{3}x}=7 => x = 81, \frac{1}{\sqrt{3}} {/tex}
But {tex} \frac{1}{\sqrt{3}} {/tex} being less than 1 is not valid
Hence, {tex}x{/tex} = 2, 81

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Q 10. The least value of n in order that the sum of first n terms of an infinite series {tex} 1 + \frac{3}{4} + \left(\frac{3}{4}\right)^{2} + \left(\frac{3}{4}\right)^{3}+ .... {/tex} should differ from the sum of the series by less then {tex} 10^{-6} {/tex} is[{tex} \ Given\ log_{10}2=0.30103, log_{10}3=0.47712 {/tex}]

A

14

B

27

53

D

57

Explanation

Series {tex} 1+\frac{3}{4}+ \left(\frac{3}{4}\right)^{2}+...+nterms=\frac{1- \left(\frac{3}{4}\right)^{n} }{1-\frac{3}{4}} {/tex}

and {tex} 1+\frac{3}{4}+ \left(\frac{3}{4}\right)^{2}+...=\frac{1}{1-\frac{3}{4}}=4 {/tex}

Now, {tex} \frac{1- \left(\frac{3}{4}\right)^{n} }{\frac{1}{4}}=4-10^{-6} {/tex}

=> {tex} 1- \left(\frac{3}{4}\right)^{n}=1-\frac{1}{4} \left(10^{-6}\right) {/tex}

=> {tex} \left(\frac{3}{4}\right)^{n}=\frac{10^{-6}}{4} {/tex}

=> {tex} nlog_{10}\frac{3}{4}=log_{10}10^{-6}-log_{10}4 {/tex}

=> {tex} n \left(0.47712-2 \times 0.30103\right)=-6-2 \times \left(0.30103\right) {/tex}

=> {tex} n=\frac{6.6026}{0.12494}=53 {/tex}

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Q 11. The identity {tex} log_{a}n\ log_{b}n\ +\ log_{b}n\ log_{c}n\ +\ log_{c}n\ log_{a}n {/tex} ls:

{tex} \frac{log_{a}n\ log_{b}n\ log_{c}n}{log_{abc}n} {/tex}

B

{tex} \frac{log_{abc}n}{log_{a}n} {/tex}

C

{tex} \frac{log_{b}n}{log_{abc}n} {/tex}

D

none of these

Explanation

{tex} log_{a}n \ log_{b}n+log_{b}n \ log_{c}n+log_{c}n \ log_{a}n {/tex}

={tex} \frac{1}{log_{n}a \ log_{n}b}+\frac{1}{log_{n}b \ log_{n}c}+\frac{1}{log_{n}c \ log_{n}a} {/tex}

{tex} \left[ \therefore log_{m}n=\frac{1}{log_{n}m} \right] {/tex}

= {tex} \frac{log_{n}c+log_{n}a+log_{n}b}{log_{n}a \ log_{n}b \ log_{n}c} {/tex}

= {tex} \frac{log_{n} \left(abc\right) }{log_{n}a \ log_{n}b \ log_{n}c}=\frac{log_{a}n \ log_{b}n \ log_{c}n}{log_{abc}n} {/tex}

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Q 12. {tex} log_{10}1000 {/tex} =

3

B

6

C

8

D

10

Explanation

{tex} log_{10}1000 {/tex} =

= {tex} log_{10}10^{3} {/tex}

={tex} 3log_{10}10 = 3 \times 1 = 3 {/tex}

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Q 13. {tex} log_{10}2500 - log_{10}25 {/tex}

A

4

B

5

2

D

1

Explanation

{tex} log_{10}2500 - log_{10}25 {/tex}

={tex} log_{10} \left(\frac{2500}{25}\right) {/tex}

={tex} log_{10}100 {/tex}

={tex} log_{10}10^{2} = 2log_{10}10 = 2 {/tex}

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Q 14. Write in terms of indices: {tex} log_{27}81=\frac{4}{3} {/tex}

{tex} 3^{4} {/tex}

B

{tex} 3^{3} {/tex}

C

{tex} 27^{\frac{1}{3}} {/tex}

D

{tex} 27^{\frac{2}{3}} {/tex}

Explanation

{tex} log_{27}81=\frac{4}{3} {/tex}

{tex} 81= \left(27\right)^{\frac{4}{3}}= \left(3^{3}\right)^{\frac{4}{3}} {/tex}

= {tex} 3^{4} {/tex}

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Q 15. Find the value of {tex} log_{3}72-log_{3}8 {/tex}

2

B

9

C

8

D

24

Explanation

{tex} log_{3}72-log_{3}8 = log_{3} \left(9 \times 8\right)-log_{3}8 {/tex}

={tex} log_{3}9+log_{3}8-log_{3}8 {/tex}

={tex} log_{3}9 {/tex}

={tex} log_{3}3^{2}=2log_{3}3 {/tex}

=2

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Q 16. Find the value of {tex} 3log_{a}2-log_{a}8 {/tex}

0

B

8

C

4

D

10

Explanation

{tex} 3log_{a}2-log_{a}8=log_{a}2^{3}-log_{a}8 {/tex}

={tex} log_{a}2^{3}-log_{a}8 {/tex}

={tex} log_{a}8-log_{a}8=0 {/tex}