Logarithm
In logarithm we will practice different types of questions on how to solve logarithmic functions on log. Solved examples on logarithm will help us to understand each and every log rules and their applications. Solving logarithmic equation are explained here in details so that student can understand where it is necessary to use logarithm properties like product rule, quotient rule, power rule and base change rule.
Click Here to understand the basic concepts on log rules.
Step-by-step solved example in Log:
1. Find the logarithms of:
(i) 1728 to the base 2√3
Solution:
Let x denote the required logarithm.
Therefore, log
_{2√3 } 1728 = x
or, (2√3)
^{x} = 1728 = 2
^{6} ∙ 3
^{3} = 2
^{6} ∙ (√3)
^{6}
or, (2√3)
^{x} = (2√3)
^{6}
Therefore, x = 6.
(ii) 0.000001 to the base 0.01.
Solution:
Let y be the required logarithm.
Therefore, log
_{0.01} 0.000001 = y
or, (0.01
^{y} = 0.000001 = (0.01)
^{3}
Therefore, y = 3.
2. Proof that, log_{2} log_{2} log_{2} 16 = 1.
Solution:
L. H. S. = log
_{2} log
_{2} log
_{2} 2
^{4}
= log
_{2} log
_{2} 4 log
_{2} 2
= log
_{2} log
_{2} 2
^{2} [since log_{2} 2 = 1]
= log
_{2} 2 log
_{2} 2
= 1 ∙ 1
= 1.
Proved.
3. If logarithm of 5832 be 6, find the base.
Solution:
Let x be the required base.
Therefore, log
_{x} 5832 = 6
or, x
^{6} = 5832 = 3
^{6} ∙ 2
^{3} = 3
^{6} ∙ (√2)
^{6} = (3 √2)
^{6}
Therefore, x = 3√2
Therefore, the required base is 3√2
4. If 3 + log_{10} x = 2 log_{10} y, find x in terms of y.
Solution:
3 + log
_{10} x = 2 log
_{10} y
or, 3 log
_{10} 10 + log
_{10} x= 1og
_{10} y
^{2} [since log_{10} 10 = 1]
or. log
_{10} 10
^{3} + log
_{10} x = log
_{10} y
^{2}
or, log
_{10} (10
^{3} ∙ x) = log
_{10} y
^{2}
or, 10
^{3} x = y
^{2}
or, x = y
^{2}/1000, which gives x in terms y.
5. Prove that, 7 log (10/9) + 3 log (81/80) = 2log (25/24) + log 2.
Solution:
Since,7 log (10/9) + 3 log (81/80) - 2 log (25/24)
= 7(log 10 – log 9)+ 3(1og 81 - log 80)- 2(1og 25 - 1og 24)
= 7[log(2 ∙ 5) - log3
^{2}] + 3[1og3
^{4} - log(5 ∙ 2
^{4})] - 2[log5
^{2} - log(3 ∙ 2
^{3})]
= 7[log 2 + log 5 – 2 log 3] + 3[4 log 3 - log 5 - 4 log 2] - 2[2 log 5 – log 3 – 3 log 2]
= 7 log 2+ 7 log 5 - 14 log 3 + 12 log 3 – 3 log 5 – 12 log 2 – 4 log 5 + 2 log 3 + 6 log 2
= 13 log 2 – 12 log 2 + 7 log 5 – 7 log 5 – 14 log 3 + 14 log 3 = log 2
Therefore 7 log(10/9) +3 log (81/80) = 2 log (25/24) + log 2.
Proved.
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6. If log_{10} 2 = 0.30103, log_{10} 3 = 0.47712 and log_{10} 7 = 0.84510, find the values of
(i) log
_{10} 45
(ii) log
_{10} 105.
(i) log_{10} 45
Solution:
log
_{10} 45
= log
_{10} (5 × 9)
= log
_{10} 5 + log
_{10} 9
= log
_{10} (10/2) + log
_{10} 3
^{2}
= log
_{10} 10 - log
_{10} 2 + 2 log
_{10} 3
= 1 - 0.30103 + 2 × 0.47712
= 1.65321.
(ii) log
_{10} 105
Solution:
log
_{10} 105
= log
_{10} (7 x 5 x 3)
= log
_{10} 7 + log
_{10} 5 + log
_{10} 3
= log
_{10} 7 + log
_{10} 10/2 + log
_{10} 3
= log
_{10} 7 + log
_{10} 10 - log
_{10} 2 + log
_{10} 3
= 0.845l0 + 1 - 0.30103 + 0.47712
= 2.02119.
7. Prove that, log_{b} a × log_{c} b × log_{d} c = log_{d} a.
Solution:
L. H. S. = log
_{b} a × log
_{c} b × log
_{d} C
= log
_{c} a × log
_{d} c [since log
_{b} M × log
_{a} b = log
_{a} M]
= log
_{d} a. (using the same formula)
Alternative Method:
Let, log
_{b} a = x Since, b
^{x} = a,
log
_{c} b = y Therefore, c
^{y} = b
and log
_{d} c = z Therefore, d
^{z} = c.
Now, a = b
^{x} = (c
^{y})
^{x} = c
^{xy} = (d
^{z})
^{xy} = d
^{xyz}
Therefore log
_{d} a = xyz = log
_{b} a × log
_{c} b × log
_{d} c. (putting the value of x, y, z)
8. Show that, log_{4} 2 × log_{2} 3= log_{4} 5 × log_{5} 3.
Solution:
L. H. S. = log
_{4} 2 × log
_{2} 3
= log
_{4} 3
= log
_{5} 3 × log
_{4} 5.
Proved.
9. Show that, log_{2} 10 - log_{8} 125 = 1.
Solution:
We have, log
_{8} 125 = log
_{8} 5
^{3} = 3 log
_{8} 5
= 3 ∙ (1/log
_{5} 8) = 3 ∙ (1/log
_{5} 2
^{3}) = 3 ∙ (1/3 log
_{5} 2) = log
_{2} 5
Therefore, L.H. S. = log
^{2} 10 - log
^{8} 125 = log
_{2} 10 - log
_{2} 5
= log
_{2} (10/5) = log
_{2} 2 = 1.
Proved.
10. If log x/(y - z) = log y/(z - x) = log z/(x – y)
show that, x^{x} y^{y} z^{z} = 1
Solution:
Let, log x/(y - z) = log y/(z - x) = log z/(x – y) = k
Therefore, log x = k(y - z) ⇔ x log x = kx(y - z )
or, log x
^{x} = kx(y - z) ... (1)
Similarly, log y
^{y} = ky (z - x) ... (2)
and log z
^{z} = kz(x - y) ... (3)
Now, adding (1), (2) and (3) we get,
log x
^{x} + log y
^{y} + log z
^{z} = k (xy - xz + yz - xy + zx - yz)
or, log (x
^{x} y
^{y} z
^{z}) = k × 0 = 0 = log 1
Therefore, x
^{x} y
^{y} z
^{z} = 1
Proved.
11. If a
^{2 - x} ∙ b
^{5x} = a
^{x + 3} ∙ b
^{3x} show that,
x log (b/a) = (1/2) log a.
Solution:
a
^{2 - x} ∙ b
^{5x} = a
^{x + 3} ∙ b
^{3x}
Therefore, b
^{5x}/b
^{3x} = a
^{x + 3}/a
^{2 - x}
or, b
^{5x - 3x} = a
^{x + 3 – 2 + x}
or, b
^{2x} = a
^{2x + 1} or, b
^{2x} =a
^{2x} ∙ a
or, (b/a)
^{2x} = a
or, log (b/a)
^{2x} = log a (taking logarithm both sides)
or, 2x log (b/a) =log a
or, x log (b/a) = (1/2) log a
Proved.
12. Show that, a
^{loga2 x} × b
^{log b2 y} × c
^{log c2 z} = √xyz
Solution:
Let, p = a
^{log a2 x}
Now, taking logarithm to the base a of both sides we get,
log
_{a} p = log
_{a} a
^{log a2 x}
⇒ log
_{a} p = log
_{a2} x ∙ log
_{a} a
⇒ log
_{a} p = log
_{a2} x [since, log
_{a} a = 1]
⇒ log
_{a} p = 1/(log
_{x} a
^{2}) [since, log
_{n} m = 1/(log
_{m} n)]
⇒ log
_{a} p = 1/(2 log
_{x} a)
⇒ log
_{a} p = (1/2) log
_{a} x
⇒ log
_{a} p = log
_{a} x
^{ ½ }
⇒ log
_{a} p = log
_{a} √x
Therefore, p = √x or, a
^{loga2 x} = √x
Similarly, b
^{logb2 y} = √y and c
^{logc2 z} = √z
L.H.S = √x ∙ √y ∙ √z = √xyz
Proved.
13. If y = a
^{1/(1 – loga x)} and z = a
^{1/(1 – loga y)} show that, x = a
^{1/(1 – loga z)}
Solution:
Let, log
_{a} x = p, log
_{a} y = q and log
_{a} z = r
Then, by problem, y = a
^{1/(1 - p)} ...……….. (1)
and z = a
^{1/(1 - q)} .............. (2)
Now, taking logarithm to the base a of both sides of (1) we get,
log
_{a} y = log
_{a} a
^{1/(1 - p)}
or, q = 1/(1 – p), [since log
_{a} a = 1]
Again, taking logarithm to the base a of both sides of (2) we get,
log
_{a} z = log
_{a} a
^{1/(1 - q)}
or, r = 1/(1 – q)
or, 1 - q = 1/r
or, 1 - 1/(1 – p) = 1/r
or, 1 - 1/r = 1/(1 – p)
or, (r – 1)/r = 1/(1 – p)
or, 1 - p = r/(r – 1)
or, p = 1- r/(r – 1) = 1/(1 – r)
or, log
_{a} x = 1/(1-log
_{a} z)
or, x = a
^{1/(1 – loga z)} Proved.
14. If x, y,z are in G. P., prove that, log
_{a} x+ log
_{a} z = 2/(log
_{y} a )[x, y, z, a > 0).
Solution:
By problem, x, y, z are in G. P.
Therefore, y/x = z/y or, zx = y
^{2}
Now, taking logarithm to the base a (> 0) of both sides we get,
log
_{a} zx = log
_{a} y
^{2} [since x, y, z > 0]
or, log
_{a} x + log
_{a} z = 2 log
_{a} y
= 2/(log
_{y} a) [since log
_{a} y × log
_{y} a = 1]
Proved.
15. Solve log
_{x} 2 ∙ log
_{x/16} 2 = log
_{x/64} 2.
Solution:
Let, log
_{2} x = a ; then, log
_{x} 2 =
1/ (log
_{2 } x) = 1/a and
log
_{x/16} 2 = 1/
[log
_{2} (x/16)] = 1/(log
_{2} x — log
_{2} 16) = 1/(log
_{2} x — log
_{2} 24)
= 1/(a – 4) [since, log
_{2} 2 = 1]
Similarly, log
_{x/64} 2 = 1/[log
_{2} (x/64)] = 1/(log
_{2} x – log
_{2} 64)
= 1/(a - log
_{2} 2
^{6}) = 1/(a – 6)
Therefore, the given equation becomes,
1/a ∙ 1/(a - 4) = 1/(a – 6)
or, a
^{2} - 4a = a - 6
or, a
^{2} - 5a + 6 = 0
or, a
^{2} - 2a - 3a + 6 = 0
or, a(a - 2) - 3(a - 2) = 0
or, (a - 2)(a - 3) = 0
Therefore, either, a - 2 = 0 i.e., a = 2
or, a - 3 = 0 i.e., a = 3
When a = 2 then, log
_{2} x = 2
therefore, x = 2
^{2} = 4
Again, when a = 3 then, log
_{2} x = 3 ,
therefore x = 2
^{3 } = 8
Therefore the required solutions are x = 4, x = 8.
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● Mathematics Logarithm
Mathematics Logarithms
Convert Exponentials and Logarithms
Logarithm Rules or Log Rules
Solved Problems on Logarithm
Common Logarithm and Natural Logarithm
Antilogarithm
11 and 12 Grade Math
Logarithms
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