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.

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, log2√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, log0.01 0.000001 = y

or, (0.01y = 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, logx 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, logb a × logc b × logd c = logd a.

Solution:

L. H. S. = logb a × logc b × logd C

= logc a × logd c     [since logb M × loga b = loga M]

= logd a. (using the same formula)

Alternative Method:

Let, logb a = x     Since, bx = a,

logc b = y     Therefore, cy = b

and logd c = z     Therefore, dz = c.

Now, a = bx = (cy)x = cxy = (dz)xy = dxyz

Therefore logd a = xyz = logb a × logc b × logd 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, xx yy zz = 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 xx = kx(y - z)         ... (1)

Similarly, log yy = ky (z - x)        ... (2)

and log zz = kz(x - y)        ... (3)

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

log xx + log yy + log zz = k (xy - xz + yz - xy + zx - yz)

or, log (xx yy zz) = k × 0 = 0 = log 1

Therefore, xx yy zz = 1     Proved.

11. If a2 - x ∙ b5x = ax + 3 ∙ b3x show that, x log (b/a) = (1/2) log a.

Solution:

a2 - x ∙ b5x = ax + 3 ∙ b3x

Therefore, b5x/b 3x = ax + 3/a 2 - x

or, b5x - 3x = ax + 3 – 2 + x

or, b 2x = a2x + 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, aloga2 x × blog b2 y × clog c2 z = √xyz

Solution:

Let, p = alog a2 x

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

loga p = loga alog a2 x

⇒ loga p = loga2 x ∙ loga a

⇒ loga p = loga2 x     [since, loga a = 1]

⇒ loga p = 1/(logx a2)     [since, logn m = 1/(logm n)]

⇒ loga p = 1/(2 logx a)

⇒ loga p = (1/2) loga x

⇒ loga p = loga x ½

⇒ loga p = loga √x

Therefore, p = √x or, aloga2 x = √x

Similarly, blogb2 y = √y and clogc2 z = √z

L.H.S = √x ∙ √y ∙ √z = √xyz     Proved.

13. If y = a1/(1 – loga x) and z = a1/(1 – loga y) show that, x = a1/(1 – loga z)

Solution:

Let, loga x = p, loga y = q and loga z = r

Then, by problem, y = a1/(1 - p)    ...……….. (1)

and z = a1/(1 - q)     .............. (2)

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

loga y = loga a1/(1 - p)

or, q = 1/(1 – p),     [since loga a = 1]

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

loga z = loga a1/(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, loga x = 1/(1-loga z)

or, x = a1/(1 – loga z)     Proved.

14. If x, y,z are in G. P., prove that, loga x+ loga z = 2/(logy 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,

loga zx = loga y2 [since x, y, z > 0]

or, loga x + loga z = 2 loga y

= 2/(logy a)   [since loga y × logy a = 1]   Proved.

15. Solve logx 2 ∙ logx/16 2 = logx/64 2.

Solution:

Let, log2 x = a ; then, logx 2 = 1/ (log2 x) = 1/a and

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

= 1/(a – 4)     [since, log2 2 = 1]

Similarly, logx/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.

Mathematics Logarithm

Mathematics Logarithms

Convert Exponentials and Logarithms

Logarithm Rules or Log Rules

Solved Problems on Logarithm

Common Logarithm and Natural Logarithm

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

Recent Articles

1. Fraction as a Part of Collection | Pictures of Fraction | Fractional

Feb 24, 24 04:33 PM

How to find fraction as a part of collection? Let there be 14 rectangles forming a box or rectangle. Thus, it can be said that there is a collection of 14 rectangles, 2 rectangles in each row. If it i…

2. Fraction of a Whole Numbers | Fractional Number |Examples with Picture

Feb 24, 24 04:11 PM

Fraction of a whole numbers are explained here with 4 following examples. There are three shapes: (a) circle-shape (b) rectangle-shape and (c) square-shape. Each one is divided into 4 equal parts. One…

3. Identification of the Parts of a Fraction | Fractional Numbers | Parts

Feb 24, 24 04:10 PM

We will discuss here about the identification of the parts of a fraction. We know fraction means part of something. Fraction tells us, into how many parts a whole has been

4. Numerator and Denominator of a Fraction | Numerator of the Fraction

Feb 24, 24 04:09 PM

What are the numerator and denominator of a fraction? We have already learnt that a fraction is written with two numbers arranged one over the other and separated by a line.