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.

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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 = 2⁶ ∙ 3³ = 2⁶ ∙ (√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, log₂ log₂ log₂ 16 = 1.

Solution:

L. H. S. = log₂ log₂ log₂ 2⁴

= log₂ log₂ 4 log₂ 2

= log₂ log₂ 2² [since log₂ 2 = 1]

= log₂ 2 log₂ 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, x⁶ = 5832 = 3⁶ ∙ 2³ = 3⁶ ∙ (√2)⁶ = (3 √2)⁶

Therefore, x = 3√2

Therefore, the required base is 3√2

4. If 3 + log₁₀ x = 2 log₁₀ y, find x in terms of y.

Solution:

3 + log₁₀ x = 2 log₁₀ y

or, 3 log₁₀ 10 + log₁₀ x= 1og₁₀ y² [since log₁₀ 10 = 1]

or. log₁₀ 10³ + log₁₀ x = log₁₀ y²

or, log₁₀ (10³ ∙ x) = log₁₀ y²

or, 10³ x = y²

or, x = y²/1000, 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) - log3²] + 3[1og3⁴ - log(5 ∙ 2⁴)] - 2[log5² - log(3 ∙ 2³)]

= 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.

6. If log₁₀ 2 = 0.30103, log₁₀ 3 = 0.47712 and log₁₀ 7 = 0.84510, find the values of

(i) log₁₀ 45

(ii) log₁₀ 105.

(i) log₁₀ 45

Solution:

log₁₀ 45 = log₁₀ (5 × 9)

= log₁₀ 5 + log₁₀ 9

= log₁₀ (10/2) + log₁₀ 3²

= log₁₀ 10 - log₁₀ 2 + 2 log₁₀ 3

= 1 - 0.30103 + 2 × 0.47712

= 1.65321.

(ii) log₁₀ 105

Solution:

log₁₀ 105

= log₁₀ (7 x 5 x 3)

= log₁₀ 7 + log₁₀ 5 + log₁₀ 3

= log₁₀ 7 + log₁₀ 10/2 + log₁₀ 3

= log₁₀ 7 + log₁₀ 10 - log₁₀ 2 + log₁₀ 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, log₄ 2 × log₂ 3= log₄ 5 × log₅ 3.

Solution:

L. H. S. = log₄ 2 × log₂ 3

= log₄ 3

= log₅ 3 × log₄ 5. Proved.

9. Show that, log₂ 10 - log₈ 125 = 1.

Solution:

We have, log₈ 125 = log₈ 5³ = 3 log₈ 5

= 3 ∙ (1/log₅ 8) = 3 ∙ (1/log₅ 2³) = 3 ∙ (1/3 log₅ 2) = log₂ 5

Therefore, L.H. S. = log² 10 - log⁸ 125 = log₂ 10 - log₂ 5

= log₂ (10/5) = log₂ 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^{log_a² 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 a²) [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^{log_a² x} = √x

Similarly, b^{log_b² y} = √y and c^{log_c² z} = √z

L.H.S = √x ∙ √y ∙ √z = √xyz Proved.

13. If y = a^{1/(1 – log_a x)} and z = a^{1/(1 – log_a y)} show that, x = a^{1/(1 – log_a 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 – log_a 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 = y²

Now, taking logarithm to the base a (> 0) of both sides we get,

log_a zx = log_a y² [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, log₂ x = a ; then, log_x 2 = 1/ (log₂ x) = 1/a and

log_x/16 2 = 1/ [log₂ (x/16)] = 1/(log₂ x — log₂ 16) = 1/(log₂ x — log₂ 24)

= 1/(a – 4) [since, log₂ 2 = 1]

Similarly, log_x/64 2 = 1/[log₂ (x/64)] = 1/(log₂ x – log₂ 64)

= 1/(a - log₂ 2⁶) = 1/(a – 6)

Therefore, the given equation becomes,

1/a ∙ 1/(a - 4) = 1/(a – 6)

or, a² - 4a = a - 6

or, a² - 5a + 6 = 0

or, a² - 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, log₂ x = 2 therefore, x = 2² = 4

Again, when a = 3 then, log₂ x = 3 , therefore x = 2³ = 8

Therefore the required solutions are x = 4, x = 8.

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