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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(i) 1728 to the base 2√3

Let x denote the required logarithm.

Therefore, log

or, (2√3)

or, (2√3)

Therefore, x = 6.

(ii) 0.000001 to the base 0.01.

Let y be the required logarithm.

Therefore, log

or, (0.01

Therefore, y = 3.

L. H. S. = log

= log

= log

= log

= 1 ∙ 1

= 1.

Let x be the required base.

Therefore, log

or, x

Therefore, x = 3√2

Therefore, the required base is 3√2

3 + log

or, 3 log

or. log

or, log

or, 10

or, x = y

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

= 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.

(i) log

(ii) log

log

= log

= log

= log

= 1 - 0.30103 + 2 × 0.47712

= 1.65321.

(ii) log

log

= log

= log

= log

= log

= 0.845l0 + 1 - 0.30103 + 0.47712

= 2.02119.

L. H. S. = log

= log

= log

Let, log

log

and log

Now, a = b

Therefore log

L. H. S. = log

= log

= log

We have, log

= 3 ∙ (1/log

Therefore, L.H. S. = log

= log

show that, x

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

Similarly, log y

and log z

Now, adding (1), (2) and (3) we get,

log x

or, log (x

Therefore, x

a

Therefore, b

or, b

or, b

or, (b/a)

or, log (b/a)

or, 2x log (b/a) =log a

or, x log (b/a) = (1/2) log a

Let, p = a

Now, taking logarithm to the base a of both sides we get,

log

⇒ log

⇒ log

⇒ log

⇒ log

⇒ log

⇒ log

⇒ log

Therefore, p = √x or, a

Similarly, b

L.H.S = √x ∙ √y ∙ √z = √xyz

Let, log

Then, by problem, y = a

and z = a

Now, taking logarithm to the base a of both sides of (1) we get,

log

or, q = 1/(1 – p), [since log

Again, taking logarithm to the base a of both sides of (2) we get,

log

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

or, x = a

By problem, x, y, z are in G. P.

Therefore, y/x = z/y or, zx = y

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

log

or, log

= 2/(log

Let, log

log

= 1/(a – 4) [since, log

Similarly, log

= 1/(a - log

Therefore, the given equation becomes,

1/a ∙ 1/(a - 4) = 1/(a – 6)

or, a

or, a

or, a

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

Again, when a = 3 then, log

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

**●** **Mathematics Logarithm**

**Convert Exponentials and Logarithms**

**Common Logarithm and Natural Logarithm**

**11 and 12 Grade Math**** ****Logarithms**** ****From Logarithm to HOME PAGE**

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