Practice Questions for Algebra - Logarithms

1. CRE 1 - Logarithms | Algebra - Logarithms

If $\log_{3} a$ = 5, find the value of a.

(a)
27
(b)
243
(c)
9
(d)
81

Answer: Option B

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Explanation :

Given, $\log_{3} a$ =5

∴ 3⁵ = a

⇒ a = 243.

Hence, option (b).

2. CRE 1 - Logarithms | Algebra - Logarithms

Find the value of $\log_{64} \frac{1}{4096}$

(a)
$- \frac{4}{3}$
(b)
-3
(c)
-2
(d)
$- \frac{1}{3}$

Answer: Option C

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Explanation :

Given, $\log_{64 (\frac{1}{4096})} \Rightarrow \log_{2^{6}} 2^{- 12} = - \frac{12}{6} \log_{2} 2$

= -2.

Hence, option (c).

3. CRE 1 - Logarithms | Algebra - Logarithms

log⁡ (343√7) to the base 7 is equal to

(a)
3
(b)
3/2
(c)
7/2
(d)
None of these

Answer: Option C

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Explanation :

$\log_{7} 343 \sqrt{7} = \log_{7} 7^{3} 7^{\frac{1}{2}} = \log_{7} 7^{\frac{7}{2}} = \frac{7}{2} \log_{7} 7 = \frac{7}{2} (\log_{a} a = 1)$

Hence, option (c).

4. CRE 1 - Logarithms | Algebra - Logarithms

$\log_{2} (3 + x) + \log_{7} 7 = 2^{3}$

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Answer: 125

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Explanation :

Given,

$\log_{2} (3 + x) + \log_{7} 7 = 2^{3} \Rightarrow \log_{2} (3 + x) + 1 = 8 \Rightarrow \log_{2} (3 + x) = 7 3 + x = 2^{7} = 128$

⇒ x = 125

Hence, 125

5. CRE 1 - Logarithms | Algebra - Logarithms

If $\log_{k} x \times \log_{7} k = 2$ , then find the value of x

(a)
$k^{7}$
(b)
$7 k^{2}$
(c)
343
(d)
49

Answer: Option D

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Explanation :

Given, $\log_{k} x \times \log_{7} k = 2$

we know,

$\log_{a} b = \frac{\log_{c} b}{\log_{c} a} ∴ \frac{\log x}{\log k} \times \frac{\log k}{\log 7} = 2 \Rightarrow \frac{\log x}{\log 7} = 2 \Rightarrow \log_{7} x = 2 \Rightarrow x = 7^{2} = 49$

Hence, option (d).

6. CRE 1 - Logarithms | Algebra - Logarithms

Find the value of $\frac{\log_{2 \sqrt{3}} 144}{\log_{\sqrt{8}} 8}$

(a)
1
(b)
2
(c)
3
(d)
4

Answer: Option B

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Explanation :

Given ,

$\frac{\log_{2 \sqrt{3}} 144}{\log_{\sqrt{8}} 8} N o w, \log_{2 \sqrt{3}} 144 = \log_{\sqrt{12}} 12^{2} = \frac{2}{\frac{1}{2}} \log_{12} 12 = 4 a n d \log_{\sqrt{8}} 8 = \frac{1}{\frac{1}{2}} \log_{8} 8 = 2 ∴ \frac{\log_{2 \sqrt{3}} 144}{\log_{\sqrt{8}} 8} = \frac{4}{2} = 2$

Hence, option (b).

7. CRE 1 - Logarithms | Algebra - Logarithms

Find the value of $\log_{10} 20 \times \log_{20} 50 \times \log_{50} 20 \times \log_{20} 30$ .

(a)
10/2
(b)
10
(c)
2
(d)
log 30

Answer: Option D

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Explanation :

We know $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$

Given, $\log_{10} 20 \times \log_{20} 50 \times \log_{50} 20 \times \log_{20} 30$

Converting all logs in base 10, we get:

= $\frac{\log 20}{\log 10} \times \frac{\log 50}{\log 20} \times \frac{\log 20}{\log 50} \times \frac{\log 30}{\log 20} = \frac{\log 30}{\log 10} = \log_{10} 30 = \log 30$

Hence, option (d).

8. CRE 1 - Logarithms | Algebra - Logarithms

Find the value $\log \frac{16}{27} - \log \frac{64}{81} + \log \frac{4}{3}$

(a)
0
(b)
1
(c)
3
(d)
log(3/4)

Answer: Option A

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Explanation :

Given $\log \frac{16}{27} - \log \frac{81}{64} + \log \frac{4}{3}$ = log 16 – log 27 – (log 64 - log 81) + log 4 – log3

= 2log4 – 3log3 – 3log4 + 4log3 + log4 – log3 = 0

Alternately,

$\log \frac{16}{27} - \log \frac{64}{81} + \log \frac{4}{3} = \log (\frac{16}{27} \times \frac{64}{81} \times \frac{4}{3})$ =

= log 1 = 0

Hence, option (a).

9. CRE 1 - Logarithms | Algebra - Logarithms

Evaluate $4^{3 - \log_{4} 3}$

(a)
192
(b)
64
(c)
64/3
(d)
$9 \log_{3} 4$

Answer: Option C

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Explanation :

Given :

$4^{3 - \log_{4} 3} = 4^{3} \times 4^{- \log_{4} 3}$

We know, $c^{\log_{a} b} = b^{\log_{a} c}$

∴ $4^{3} \times 4^{- \log_{4} 3} = 4^{3} \times 3^{- \log_{4} 4} = 4^{3} \times 3^{- 1}$ = 64/3

Hence, option (c).

10. CRE 1 - Logarithms | Algebra - Logarithms

The value of $[\frac{1}{\log_{x y} (x y z)} + \frac{1}{\log_{y z} (x y z)} + \frac{1}{\log_{z x} (x y z)}]$ is equal to

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Answer: 2

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Explanation :

We know , $\frac{1}{\log_{a} b} = \log_{b} a$ and

$\log_{a} m + \log_{a} n = \log_{a} m n G i v e n, [\frac{1}{\log_{x y} (x y z)} + \frac{1}{\log_{y z} (x y z)} + \frac{1}{\log_{z x} (x y z)}] = \log_{x y z} (x y) + \log_{x y z} (y z) + \log_{x y z} (z x) = \log_{x y z} (x y \times y z \times z x) = \log_{x y z} {(x y z)}^{2} = 2 \log_{x y z} (x y z) = 2 \times 1 = 2$

Hence , 2.

11. CRE 1 - Logarithms | Algebra - Logarithms

If $\log_{5} [\log_{2} (\log_{3} x)] = 1$ then x is

(a)
243
(b)
729
(c)
$3^{32}$
(d)
625

Answer: Option C

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Explanation :

$\log_{5} [\log_{2} (\log_{3} x)] = 1 \Rightarrow \log_{2} (\log_{3} x) = 5 \Rightarrow \log_{3} x = 2^{5} = 32 \Rightarrow 3^{32} = x$

Hence, option (c).

12. CRE 1 - Logarithms | Algebra - Logarithms

The value of $[3 \log (\frac{81}{80}) + 5 \log (\frac{25}{24}) + 7 \log (\frac{16}{15})]$ is

(a)
log3
(b)
log5
(c)
log7
(d)
log2

Answer: Option D

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Explanation :

Given,

$3 \log \frac{81}{80} + 5 \log \frac{25}{24} + 7 \log \frac{16}{15} = \log [{(\frac{81}{80})}^{3} \times {(\frac{25}{24})}^{5} \times {(\frac{16}{15})}^{7}] = \log (\frac{3^{12} \times 5^{10} \times 2^{28}}{2^{12} \times 5^{3} \times 2^{15} \times 3^{5} \times 3^{7} \times 5^{7}})$

= log⁡2

Hence, option (d).

13. CRE 1 - Logarithms | Algebra - Logarithms

If $\log_{10} a + \log_{10} b = c$ ,then the value of a is

(a)
bc
(b)
c/b
(c)
$\frac{{(10)}^{c}}{b}$
(d)
$\frac{10 b}{c}$

Answer: Option C

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Explanation :

Given

$\log_{10} a + \log_{10} b = c \Rightarrow \log_{10} (a b) = c \Rightarrow 10^{c} = a b \Rightarrow a = \frac{{(10)}^{c}}{b}$

Hence, option (c).

14. CRE 1 - Logarithms | Algebra - Logarithms

If $\log_{y} x = 8 a n d \log_{10 y} 16 x = 4$ , then find the value of y.

(a)
1
(b)
2
(c)
3
(d)
5

Answer: Option D

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Explanation :

Given, logx = 8

=> y⁸= x

$\log_{10 y} 16 x = 4 \Rightarrow 10^{4} y^{4} = 16 x$

Dividing (2) by (1)

$10^{4} y^{- 4} = 16$

⇒y=5

Hence, option (d).

15. CRE 1 - Logarithms | Algebra - Logarithms

log 0.0867=?

(a)
log 8.67 +2
(b)
log 8.67 - 2
(c)
$\frac{\log 867}{1000}$
(d)
-2log 8.67

16. CRE 1 - Logarithms | Algebra - Logarithms

Find x, if 0.01^x = 2

(a)
log 2/2
(b)
2/log2
(c)
-2/ log2
(d)
-log 2/2

Answer: Option D

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Explanation :

Given, 0.01^x = 2.

⇒x= $\log_{0.01} 2 = \log_{10^{- 2}} 2 = \frac{\log 2}{- 2}$

Hence, option (d).

17. CRE 1 - Logarithms | Algebra - Logarithms

If $2^{x} \times 3^{2 x} = 100$ , , then the value of 𝑥 is (log 2 = 0.3010, log 3 = 0.4771)

(a)
2.3
(b)
1.59
(c)
1.8
(d)
1.41

Answer: Option B

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Explanation :

Given, $2^{x} \times 3^{2 x} = 100 \Rightarrow x \log 2 + 2 x \log 3 = \log 100$

⇒x(0.3010 + 2 × 0.4771) = 2

$x = \frac{1}{1.2552} = 1.59$

Hence, option (b).

18. CRE 1 - Logarithms | Algebra - Logarithms

If log 2 = 0.30103, find the number of digits in 2⁶⁰

(a)
19
(b)
31
(c)
100
(d)
200

19. CRE 1 - Logarithms | Algebra - Logarithms

The mantissa of log 3274 is 0.5150. The value of log (0.3274) is

(a)
1.5150
(b)
1.5150
(c)
2.5150
(d)
None of these

20. CRE 1 - Logarithms | Algebra - Logarithms

If $\log_{10} a = b t h e n f i n d t h e v a l u e o f 10^{3 b} i n t e r m s o f a$ .

(a)
a³
(b)
3a
(c)
ax1000
(d)
ax100

CRE 1 - Logarithms | Algebra - Logarithms

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