Today our topic of discussion is Logarithms of Mathematics.
Logarithms of Mathematics
Logarithms
Logarithms are used to find the values of exponential expressions. Logarithm is written in brief as Log. Product, quotient, etc. of large numbers or quantities can easily be determined with the help of log.
We know, 23= 8 this mathematical statement is written in terms of log as log_2(8) = 3 Again, conversely, if log_2(8) = 3 it can be written in terms of exponents as 2 ^ 3 = 8 That is, if 2 ^ 3 = 8 , then log_2(8) = 3 and conversely, if 23 = 8 Similarly, 2 ^ – 3 = 1/(23) = 1/8 can be written in terms of log as, log_2(8) = 3 then log_2(1/8) = – 3
If ax = N ,(a>0,a ne1) then x = log_a(N) is defined as a based log_a(N)
Note: If a > 0 then a ^ x is always positive whatever may be the values of r, positive or negative numbers. So, only the log of positive numbers has values that are real; log of zero or negative numbers have no real value.
Laws of Logarithms
Let, a > 0 a ne1; l / b > 0 ,b ne1 a M > 0, N > 0
Formula 6 (zero and one log). If a > 0, a ne11) log_a(1) = 0 2) log_a(a) = 1
Proof: We know from the formula of exponents, a ^ 0 = 1
.. from the definition of log, we get, log_a(1) = 0 (proved)
Again, we know, from the formula of exponents, a ^ 1 = a
.. from the definition of log, we get, log_a(a) = 1 (proved)
Formula 7 (Log of products). log_a(MN) = log_a(M) + log_a(N)
Proof: Let, log_a(M) = x log_a(N) = y
* M = a ^ x N = a ^ y
Now, MN = a ^ x * a ^ y = a ^ (x + y)
log_a(MN) = x + y
or, log_a(MN) = log_a(M) + log_a(N) putting the values of r, y]
log_a(MN) = log_a(M) + log_a(N(proved))
Note: log a (MNP…)= log_a(M) + log_a(N) + log_a(P) +…
Note: log a (M pm N) ne log_a(M) plus/minus log_a(N)
Formula 8 (log of quotient). log_a(M/N) = log_a(M) – log_a(N)
Proof: Let, log_a(M) = x, log_a(N) = y
therefore M = ax ,N=ay
Now, M/N = (ax)/(ay) = a(x-y)
. log_a(M/N) = x – y
g_{a} * M/N =log a M-log a N(pr )
Formula 9 (Log of power). log_a(Mr) = r * log_a(M)
Proof: Let, log_a(M) = x ,..M=a^ x or, ,(M)r =(ax )r or, Mr = a ^ (rx)
. log_a(M r) = rxor , log_a(Mr) = r * log_a(M)
*. log_a(Mr )= r * log_a(M) (proved)
Note: (log_a(M))r and rlog, M may not be equal.
For example, (log,4)5 (log, 22)52532, 5 * log_2(4) = 5 * 2 =10 ne32
Formula 10 (change of base). log_a(M) = log_b(M) * log_a(b)
Proof: Let, log_a(M) = x, log_b(M) = y
., ax = M by = M
.. ax = by or, (ax) ^ (1/y) = (b ^ y) ^ (1/y) * or b = a ^ (pi/y)
therefore x/y = log_a(bor) ,x=ylog a b
or ,log a M=log b M* log a b(proved)
Corollary 1. log_a(b) = 1/(log_b(a)) or, log_b(a) = 1/(log_a(b))
Proof: We know, log_a(M) = log_b(M) * log_a(b)
Putting M = a we get, log_a(a) = log_b(a) * log_a(b) or, 1 = log_b(a) * log_a(b)
.. log_a(b) = 1/(log_b(a)) or log_b(a) = 1/(log_a(b)) (proved)
Example 6. Find the values:
1) log_10(100)
2) log_3(1/9)
3) log_(sqrt(3))(81)
Solution:
1) log 10 100=log 10 102 =2log 10 10[** log_10(Mr) = rlo*g_{10}*M ] =2*1=2[** log_a(a) = 1 ]
2) log_3(1/9) = log_3(1/(32)) = log_3(3-2) =-2 log 3 3[/* log a M^ r =rlog a M]
=-2*1=-2[** log_a(a) = 1 ]
3) log_(sqrt(3))(81) = log_(sqrt(3))(3 ^ 4)
= log_(sqrt(3))\{(sqrt(3)) ^ 2\} ^ 4
= log_(sqrt(3))(sqrt(3)) ^ 8
= 8 * log_(sqrt(3))(sqrt(3))
= 8 * 1 =8[** log a a=1]
Example 7.
1) What is the log of 5sqrt(5) to the base of 5? 400 is 4, then what is the base of log?
2) If the log of
Solution:
1) Log of 5sqrt(5) to the base of 5
= log_5(5) * sqrt(5) = log_5(5 * 5 1/2) = log_5(53/2)
= 3 2 log 5 5[** log_a(M ^ r) = r * log_a(M) ]
= 3/2 * 1 = 3 2 [::log a a=1]
2) Let the base be a
.. By the question, log_a(400) = 4
* , a4 = 400or ,a4 =(20)2 =\ (2 sqrt 5 )2 \2 =(2 sqrt 5 )4
therefore a = 2sqrt(5)
overline 5 [** ax = bx ,ax ne 0, a = b
.. Base 2sqrt(5)
Example 8.
Find the value of x:
1) log_10(x) = – 2
2) log_x(324) = 4
Solution:
1) log_10(x) = – 2
or, x = 10-2 = 1/(10 ^ 2) = 1/100 = 0.01 * x = 0.01
2) log_x(324) = 4
or, x ^ 4 = 324 = 3 * 3 * 3 * 3 * 2 * 2 = 34 * 2 ^ 2 sr, x4 = 3 4 * (sqrt(2))4 or, x4 = (3sqrt(2))4
x = 3√(2)
Example 9. Prove that, 3 * log_10(2) + log_10(5) = log_10(40)
Solution: Left hand side = 3 * log_10(2) + log_10(5)
= log_10(2 ^ 3) + log_10(5) [*** log_a(M ^ r) = r * log_a(M) ]
= log_10(8) + log_10(5)
= log_10(8 * 5) [** log_a(MN) = log_a(M) + log_a(N) ]
= log_10(40) = right hand side (proved)
Example 10. Simplify: (log_10(√(27)) + log_10(8) – log_10(√(1000)))/(log_10(1.2))
Solution: (log_10(√(27)) + log_10(8) – log_10(√(1000)))/(log_10(1.2))
= (log_10(3³) ^ (1/3) + log_10(8) – log_10(10 ³) ^ (1/3))/(log_10(12/10))
= (log_10(3 ^ (3/2)) + log_10(2³) – log_10(10) ^ (3/3))/(log_10(12) – log_10(10))
= (3/2 * log_10(3) + 3 * log_10(2) – 3/2 * log_10(10))/(log_10(3 * 2 ^ 2) – log_10(10))
= (3/2 * (log_10(3) + 2 * log_10(2) – 1))/(log_10(3) + 2 * log_10(2) – 1) [*** log_10(10) = 1 ]
= 3/2
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