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 1/4Solution: **

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

**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)

or, log₅ 32 = (log₁₀ 2 ⁵)/(log₁₀ 5)

= (5 log ₁₀ 2)/(log₁₀ 5)

= (5 × 0.30103)/0.69897

= 2.15 (approx.)

Solution:

log

or, 1/(log

or, 1/(log

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

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

or, 3a = 1

or, a = 1/3

or, log

or, x

or, x = 8.

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

Therefore, log (1 + ac) = log b

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

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(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.

Solution:

Let, P = (yz)

Therefore log P = log [(yz)

= log (yz)

= (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)

or, log P = 0 = log 1

P = 1 i.e., (yz)

Solution:

We have, log

Therefore, 1/(log

Similarly,1/(log

L.H.S. = 1/(log

= log

log

5

= 5

or, a

or, a

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

and, a = 25 gives, 5

Therefore, 1/(2x) = 2 or, x = 1/4

Therefore the required solutions are x = 1/2 or, x = 1/4.

prove that, log

Let,log

Then by definition of logarithm we have,

N = a

Therefore a = N

Now, b = √ac or, b

or, (N

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

● **Convert Exponentials and Logarithms**

● **Logarithm Rules or Log Rules**

● **Worked-Out Problems on Logarithm**

● **Solved Examples in Logarithm**

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