Subscribe to our YouTube channel for the latest videos, updates, and tips.


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





11 and 12 Grade Math 

From Solved Examples in Logarithm to HOME PAGE




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.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. 5th Grade Circle Worksheet | Free Worksheet with Answer |Practice Math

    Jul 10, 25 11:41 AM

    Radii of the circRadii, Chords, Diameters, Semi-circles
    In 5th Grade Circle Worksheet you will get different types of questions on parts of a circle, relation between radius and diameter, interior of a circle, exterior of a circle and construction of circl…

    Read More

  2. Construction of a Circle | Working Rules | Step-by-step Explanation |

    Jul 09, 25 01:29 AM

    Parts of a Circle
    Construction of a Circle when the length of its Radius is given. Working Rules | Step I: Open the compass such that its pointer be put on initial point (i.e. O) of ruler / scale and the pencil-end be…

    Read More

  3. Combination of Addition and Subtraction | Mixed Addition & Subtraction

    Jul 08, 25 02:32 PM

    Add and Sub
    We will discuss here about the combination of addition and subtraction. The rules which can be used to solve the sums involving addition (+) and subtraction (-) together are: I: First add

    Read More

  4. Addition & Subtraction Together |Combination of addition & subtraction

    Jul 08, 25 02:23 PM

    Addition and Subtraction Together Problem
    We will solve the different types of problems involving addition and subtraction together. To show the problem involving both addition and subtraction, we first group all the numbers with ‘+’ and…

    Read More

  5. 5th Grade Circle | Radius, Interior and Exterior of a Circle|Worksheet

    Jul 08, 25 09:55 AM

    Semi-circular Region
    A circle is the set of all those point in a plane whose distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is known

    Read More