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 log₁₀ 2 lies between 1/3 and 1/4

Solution: 


We have, 8 < 10 or, 2³ < 10 

Therefore, log₁₀ 2³ < log₁₀ 10

or, 3log₁₀ 2 < 1 

or, log₁₀ 2 < 1/3     ................... (i) 

Again, 16 > 10 or, 2⁴ > 10 

or, log₁₀ 2⁴ > log₁₀ 10 

or, 4 log₁₀ 2 > 1 

or, log₁₀ 2 > 1/4     .................. (ii) 

Therefore from (i) and (ii) we get, 1/3 > log₁₀ 2 > 1/4 

i.e., the value of log₁₀ 2 lies between 1/3 and 1/4     Proved.


2. If log₁₀ 2 = 0.30103 find the value of log₅ 32. 

Solution: 


Since, log₁₀ 10 = 1, 

Therefore, log₁₀ (2 × 5) = 1 

or, log₁₀ 2 + log₁₀ 5 = 1 

or, log₁₀ 5 = 1 - log₁₀ 2 = 1 — 0.30103 = 0.69897 

Now, log₅ 32 = log₁₀ 32 × log₅ 10 = (log₁₀ 32)/(log₁₀ 5) [since log₅ 10 × log₁₀  5 = 1] 

or, log₅ 32 = (log₁₀ 2 ⁵)/(log₁₀ 5) 

               = (5 log ₁₀ 2)/(log₁₀ 5) 

               = (5 × 0.30103)/0.69897 

               = 2.15 (approx.) 


3. If log2 x + log4 x + log16 x = 21/4 find the value of x.

Solution:


log2 x + log4 x + log16 x = 21/4

or, 1/(logx 2) + 1/(logx 4) + 1/(logx 16) = 21/4     [since, log2 x × logx 2 = 1]

or, 1/(logx 2) + 1/(2 logx 2) + 1/(4 logx 2) = 21/ 4

or, 1/a + a/(2a) + 1/(4a) = 21/4     [assuming logx 2 = a]

or, 7/(4a) = 21/4

or, 3a = 1

or, a = 1/3

or, logx 2 = 1/3

or, x1/3 = 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 + b2 - 1 = b2

Therefore, log (1 + ac) = log b2 = 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, y2 = 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 (y2) + (c - a) log y [since xz = y2]

= 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/(loga bc + 1) + 1/(logb ca + 1) + 1/(logc ab + 1) = 1.

Solution:


We have, loga bc + 1 = loga bc + loga a = loga abc

Therefore, 1/(loga bc) + 1 = 1/(loga abc) =logabc a

Similarly,1/(logb ca + 1) = logabc b and 1/(logc ab + 1) = logabc c

L.H.S. = 1/(loga bc) + 1 + 1/(logb ca) + 1 + 1/(logc ab + 1)

= logabc a + logabc b + logabc c = logabc (abc) = 1.     Proved.



8. Solve log5 (51/x + 125) = log5 6 + 1 + 1/(2x)

Solution:

log5 (51/x + 125) = log5 6 + 1 + 1/(2x)

51/x + 125 = 5log5 6 + 1 + 1/(2x) [ since loga M = x implies ax = M]

= 5log5 6 ∙ 51 ∙ 51/(2x) = 6 ∙ 5 ∙ 51/(2x) [since aloga M = M]

or, a2 + 125 = 30a     where a = 5 1/(2x)

or, a2 - 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) = 52

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, loga N/logc N = (loga N - logb N)/(logb N - logc N)


Solution:

Let,loga N= x, logb N = y and logc N = z.

Then by definition of logarithm we have,

N = ax = by = cz

Therefore a = N1/x , b = N1/y and c = N1/z

Now, b = √ac or, b2 = ac

or, (N 1/y)2 = N1/x ∙ N1/z or, N2/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, loga N/logc N = (loga N - logb N)/(logb N - logc N) [putting the values of x, y, z]     Proved.

Convert Exponentials and Logarithms
Logarithm Rules or Log Rules
Worked-Out Problems on Logarithm
Solved Examples in Logarithm





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