Logarithm, often called ‘logs,’ is the power to which a number must be raised to get the result. It is thus the inverse of the exponent and is written as:
ba = x ⇔ logbx = a
Here,
‘b’ is the base
‘a’ is the exponent
‘x’ is the argument
are the 3 parts of a logarithm.
Thus, the logarithm represents the exponent to which a base is raised to yield a given number.
How to Undo Logarithms
For example, we know 43 = 64
Here, using the logarithm, we can answer how many 4s multiply to get 64.
Since 4 × 4 × 4 = 64, we multiply three 4s to get 64, which is written in the logarithmic form as log4(64) = 3, read as ‘log base 4 of 64 is 3.’
Thus, 43 = 64 ⇔ log4(64) = 3, where the base is 4, and the exponent or power is 3
Here are some examples of conversions from exponential to logarithmic form and vice-versa.
Exponential Form
Logarithmic Form
54 = 625
log5(625) = 4
152 = 225
log15(225) = 2
83 = 512
log8(512) = 3
Find the value of log7(343).
Solution:
As we know, 7 × 7 × 7 = 73 = 343 Thus, log7(343) = 3
Convert 35 = 243 in its logarithmic form.
Solution:
As we know, ba = x ⇒ logbx = a Here, 35 = 243 ⇒ log3(243) = 5, the required logarithmic form.
However, the expression logbx has some restrictions, which are as follows:
Restriction on the Variable
For Base
The base ‘b’ of a logarithm is always a positive real number (b > 0) and does not equal 1 (b ≠ 1).
For negative bases, logarithm leads to complex results. Now, let us assume the base is 1, and the equation is:
log17 = x ⇒ 1x = 7
Since 1 raised to any power yields 1, 1x = 7 is false. Thus, the base does not equal 1.
For Argument
The argument ‘x’ is always a positive real number (a > 0).
Since a positive number (b > 0) is raised to any power, it yields a positive number (bx > 0).
Thus, bx = a follows that a > 0.
Common Logarithms (log) vs Natural Logarithm (ln)
The base of a logarithm can have many positive values except 1. However, two of them are frequently used.
Common Logarithm
The logarithm whose base is 10 is known as the common logarithm or base-10-logarithm. It is often denoted as log(x) without a subscript.
For example, log10(10000) = log(10000) = 4 ⇔ 104 = 10000
Natural Logarithm
The logarithm with the base ‘e’ (≈ 2.718…, Euler’s number) is the natural logarithm or base-e-logarithm, denoted by ln(x) or loge(x).
For example, ln(e2) = 2 ⇔ e2 = e × e, ln(9) = c ⇔ ec = 9
Rules
Certain rules (also properties or identities) of logarithms are used to simplify, expand, or condense them.
Simplifying
Let us simplify a logarithmic function log35 using the appropriate rule(s).
Since logb(xy) = logbx + logby and 35 = 5 × 7 (Using Product rule)
Here, ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$ = ${\log \left( \dfrac{10x^{4}}{y^{2}}\right) +2\log \left( y\right)}$ Using the quotient rule, we get ${\log \left( 10x^{4}\right) -\log \left( y^{2}\right) +2\log \left( y\right)}$ Using the product rule, we get ${\log \left( 10\right) +\log \left( x^{4}\right) -\log \left( y^{2}\right) +2\log \left( y\right)}$ Using the power rule, we get ${\log \left( 10\right) +4\log \left( x\right) -2\log \left( y\right) +2\log \left( y\right)}$ Since logbb = 1, log(10) = 1 Thus, ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$ = 1 + 4log(x)
Expand the expression log2(16x8)
Solution:
Here, log2(16x8) Using the product law of logarithms, we get log2(16) + log2(x8) = 4 + log2(x8) Using the power law of logarithms, we get 4 + 8 log2(x) Thus, log2(16x8) = 4 + 8 log2(x)
Condense the expression log5(9x2) – log5(x) + log5(8x4)
Solution:
Here, log5(9x2) – log5(x) + log5(8x4) First, we use the quotient rule by subtracting the logarithms. The expression is: ${\log _{5}\left( \dfrac{9x^{2}}{x}\right) +\log _{5}\left( 8x^{4}\right)}$ = ${\log _{5}\left( 9x\right) +\log _{5}\left( 8x^{4}\right)}$ Now, we use the product rule by adding the logarithms. The expression becomes ${\log _{5}\left( 9x\cdot 8x^{4}\right)}$ = ${\log _{5}\left( 72x^{5}\right)}$ Thus, log5(9x2) – log5(x) + log5(8x4) = ${\log _{5}\left( 72x^{5}\right)}$
Solve the equation log(x) + log(3) = log(51) and find the value of x.
Solution:
Here, log(x) + log(3) = log(51) ⇒ log(3x) = log(51) (by the product rule) Now, canceling the log from both sides, we get 3x = 51 ⇒ x = 17 Thus, x = 17
Solving Nested Logarithm
A nested logarithm, also called iterated logarithm or repeated logarithm, represents a logarithm within another logarithm and is denoted as log(log(log…(logx))) or logn(x), where ‘n’ is the level of nesting.
Let us solve the nested logarithm log2(log8(64))
First, we solve the inner logarithm log8(64). Since 82 = 64, log8(64) = 2
Now, solving the outer logarithm, we get log2(2) = 1
Thus, log2(log8(64)) is simplified to 1.
Simplify log5(log3(log7(343)))
Solution:
Here, log5(log3(log7(343))) Since 73 = 343, log7(343) = 3 Solving the inner logarithm, we get log5(log3(3)) Since logb(b) = 1, log3(3) = 1 Now, solving further logarithms, we get log5(1) = 0
