Solved Examples in Logarithm
Solved examples in logarithm are explained here step by steep using different laws of logarithm. Some worked-out problems on logarithm will help us with the concept how to apply all the formulas of logarithm.
1. Show that the value of log_{10} 2 lies between 1/3 and 1/4
Solution:
We have, 8 < 10 or, 2^{3} < 10
Therefore, log_{10} 2^{3} < log_{10} 10
or, 3log_{10} 2 < 1
or, log_{10} 2 < 1/3 ................... (i)
Again, 16 > 10 or, 2^{4} > 10
or, log_{10} 2^{4} > log_{10} 10
or, 4 log_{10} 2 > 1
or, log_{10} 2 > 1/4 .................. (ii)
Therefore from (i) and (ii) we get, 1/3 > log_{10} 2 > 1/4
i.e., the value of log_{10} 2 lies between 1/3 and 1/4 Proved.
2. If log_{10} 2 = 0.30103 find the value of log_{5} 32.
Solution:
Since, log_{10} 10 = 1,
Therefore, log_{10} (2 × 5) = 1
or, log_{10} 2 + log_{10} 5 = 1
or, log_{10} 5 = 1 - log_{10} 2 = 1 — 0.30103 = 0.69897
Now, log_{5} 32 = log_{10} 32 × log_{5} 10 = (log_{10} 32)/(log_{10} 5) [since log_{5} 10 × log_{10} 5 = 1]
or, log_{5} 32 = (log_{10} 2 ^{5})/(log_{10} 5)
= (5 log _{10} 2)/(log_{10} 5)
= (5 × 0.30103)/0.69897
= 2.15 (approx.)
3. If log_{2} x + log_{4} x + log_{16} x = 21/4 find the value of x.
Solution:
log_{2} x + log_{4} x + log_{16} x = 21/4
or, 1/(log_{x} 2) + 1/(log_{x} 4) + 1/(log_{x} 16) = 21/4 [since, log_{2} x × log_{x} 2 = 1]
or, 1/(log_{x} 2) + 1/(2 log_{x} 2) + 1/(4 log_{x} 2) = 21/ 4
or, 1/a + a/(2a) + 1/(4a) = 21/4 [assuming log_{x} 2 = a]
or, 7/(4a) = 21/4
or, 3a = 1
or, a = 1/3
or, log_{x} 2 = 1/3
or, x^{1/3} = 2
or, x = 8.
4. If a, b, c are three consecutive positive integers, show that, log (1 + ac) = 2 log b.
Solution:
Since, a, b, c are three consecutive positive integers, hence, either
a = b – 1 and c = b + 1
or, a = b + 1 and c = b - 1
Clearly, in any case, 1 + ac = 1 + (b - 1)(b + 1) = 1 + b^{2} - 1 = b^{2}
Therefore, log (1 + ac) = log b^{2} = 2 log b. Proved.
5. If a, b, c are in A. P. and x, y, z are in G. P., prove that, (b - c) log x + (c - a) log y + (a - b) log z = 0.
Solution:
By problem, a, b and c are in A. P.
Therefore b - a = c - b or, a – b = b - c
or, 2b – c – a = 0
Again, x, y, z are in G. P.
Therefore y/x = z/y or, y^{2} = zx.
L. H. S. = (b - c) log x + (c - a) log y + (a - b) log z
= (b - c) log x + (c - a) log y + (b - c) log z [since a - b = b - c]
= (b - c) (log x + log z)+ (c - a) log y
= (b - c) log (xz) + (c - a) log y
= (b - c) log (y^{2}) + (c - a) log y [since xz = y^{2}]
= 2(b - c) log y + (c - a) log y
= log y × (2b - 2c + c - a)
= log y × (2b - c - a)
= log y × 0 [since 2b – c – a = 0]
= 0. Proved.
6. Prove that, (yz)^{log (y/z)} × (zx)^{log (z/x)} × (xy)^{log (x/y)} = 1.
Solution:
Let, P = (yz)^{log (y/z)} × (zx)^{log (z/x)} × (xy)^{log (x/y)}
Therefore log P = log [(yz)^{log (y/z)} × (zx)^{log (z/x)} × (xy)^{log (x/y)}]
= log (yz)^{log (y/z)} × log (zx)^{log (z/x)} × log (xy)^{log (x/y)}
= (log y - log z)(log y + log z) + (log z - log x)(log z + log x) + (log x - log y)(log x + logy)
= (log y)^{2} - (log z)^{2} + (log z)^{2} - (log x)^{2} + (log x)^{2} (log y)^{2}
or, log P = 0 = log 1
P = 1 i.e., (yz)^{log (y/z)} × (zx)^{log (z/x)} × (xy)^{log (x/y)} = 1.
7. Show that, 1/(log_{a} bc + 1) + 1/(log_{b} ca + 1) + 1/(log_{c} ab + 1) = 1.
Solution:
We have, log_{a} bc + 1 = log_{a} bc + log_{a} a = log_{a} abc
Therefore, 1/(log_{a} bc) + 1 = 1/(log_{a} abc) =log_{abc} a
Similarly,1/(log_{b} ca + 1) = log_{abc} b and 1/(log_{c} ab + 1) = log_{abc} c
L.H.S. = 1/(log_{a} bc) + 1 + 1/(log_{b} ca) + 1 + 1/(log_{c} ab + 1)
= log_{abc} a + log_{abc} b + log_{abc} c = log_{abc} (abc) = 1. Proved.
8. Solve log_{5} (5^{1/x} + 125) = log_{5} 6 + 1 + 1/(2x)
Solution:
log_{5} (5^{1/x} + 125) = log_{5} 6 + 1 + 1/(2x)
5^{1/x} + 125 = 5^{log5 6 + 1 + 1/(2x)} [ since log_{a} M = x implies a^{x} = M]
= 5^{log5 6} ∙ 5^{1} ∙ 5^{1/(2x)} = 6 ∙ 5 ∙ 5^{1/(2x)} [since a^{loga M} = M]
or, a^{2} + 125 = 30a where a = 5 ^{1/(2x)}
or, a^{2} - 30a + 125 = 0
or, (a - 5)(a - 25) = 0
Therefore either, a - 5 = 0 i.e., a = 5
or, a - 25 = 0 i.e., a = 25
Now, a = 5 gives 5 ^{1/(2x)} = 5 Therefore, 1/(2x) = 1 or, x = 1/2
and, a = 25 gives, 5 ^{1/(2x)} = 5^{2}
Therefore, 1/(2x) = 2 or, x = 1/4
Therefore the required solutions are x = 1/2 or, x = 1/4.
9. If a > 0, c > 0, b = √aca, c and ac ≠ 1, N > 0,
prove that, log_{a} N/log_{c} N = (log_{a} N - log_{b} N)/(log_{b} N - log_{c} N)
Solution:
Let,log_{a} N= x, log_{b} N = y and log_{c} N = z.
Then by definition of logarithm we have,
N = a^{x} = b^{y} = c^{z}
Therefore a = N^{1/x} , b = N^{1/y} and c = N^{1/z}
Now, b = √ac or, b^{2} = ac
or, (N ^{1/y})^{2} = N^{1/x} ∙ N^{1/z} or, N^{2/y} = N ^{1/x + 1/z}
Therefore, 2/y = 1/x + 1/z
or, 1/y - 1/x = 1/z - 1/y
or, (x - y)/x = (y – z)/z
or, x/z = (x – y)/(y – z)
or, log_{a} N/log_{c} N = (log_{a} N - log_{b} N)/(log_{b} N - log_{c} N) [putting the values of x, y, z] Proved.
Related Links:
● Convert Exponentials and Logarithms
● Logarithm Rules or Log Rules
● Worked-Out Problems on Logarithm
● Solved Examples in Logarithm
11 and 12 Grade Math
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