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

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

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

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

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

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

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

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

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 $4^{3 - \log_{4} 2}$

A.
192
B.
64
C.
64/3
D.
$9 \log_{3} 4$

Answer: Option C

Explanation :

Given :

$4^{3 - \log_{4} 3} = 4^{3} \times 4^{- \log_{4} 3} W e k n o w, c^{\log_{a} b} = b^{\log_{a} c} ∴ 4^{3} \times 4^{- \log_{4} 3} = 4^{3} \times 3^{- \log_{4} 4} = 4^{3} \times 3^{- 1} = 64 / 3$

Hence, option (c).

If log2 [log3 (log2x)]=0, then x is equal to

Solution:

$\log_{2} [\log_{3} (\log_{2} x)] = 0 \Rightarrow \log_{3} (\log_{2} x) = 2^{0} = 1 \Rightarrow \log_{2} x = 3 \Rightarrow x = 2^{3} = 8$

Hence , 8

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

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

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

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

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

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

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

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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