If you are someone who is taking a college or high school mathematics class, you are most likely to cover the logarithms section and rules of Logarithm. But what is Logarithm? Why does this letter “e” pop up in this mathematical expression? Natural logs might seem a bit difficult. However, once you get a glimpse of the Log rules, you can easily solve the complicated logarithm problems. Here, we explain 7 important rules for Logarithm or laws of Logarithm.

The Logarithm or ln can be defined as the inverse for “e”. Here, this letter “e” represents the mathematical constant known as the natural exponent. “e” comes as a mathematical constant with a properly set value. The value carried by “e” equals 2.71828. “e” appears at several mathematical instances for Logarithm rules which include scenarios such as growth equations, compound interest, as well as decay equations. ln(x) is termed as the time required for grow into “x”. On the other hand, ex is termed as the growth which has occurred post the time “x”.

There are 7 primary rules that are required to be adhered to when
working upon the Log properties.

Here are some ways on how to
expand Logarithm.

- ln(x)(y) = ln(x) + ln(y)
- The log result for multiplication of the variables x & y is actually the sum of ln(x) and ln(y)
- For example: ln(8)(6)= ln(8) + ln (6)

- ln(x/y)= ln(x) – ln(y)
- The log result for the division of the variables x & y is actually the difference of in(x) and ln(y)
- For example: ln(7/4)= ln(7) – ln(4)

- ln(x)=0
- The log result, in this case, is equal to zero
- For example: ln(1)=0

- Inx(x)=1
- In this case, Logarithm of the number which is equal with its base is equal to 1
- For example: ln
_{3}(3)=1

- ln(1/x)= -ln(x)
- The log result for the reciprocal result of x here is the exact opposite of ln (x)
- For example: ln(1/3)= -ln(3)

- ln(xy) = y* ln(x)
- The log result here for x is raised to power by y around y times for ln(x)
- For example: ln(52)= 2* ln(5)

- lnx(xk)= k
- In this case the Logarithm for exponential number with the same base as log, the result is the exponent
- For example: ln
_{3}(32)=2

Apart from these 4 rules discussed here, there are many properties for Logarithm you need to know. Here are some popular log properties.

Scenario | (ln) Property |
---|---|

(ln) of any negative number | The (ln) for any negative number = undefined |

(ln) of 0 | ln(0) = undefined |

(ln) of 1 | ln(1) = 0 |

ln of Infinity | ln(∞) = ∞ |

(ln) of e | ln(e) = 1 |

(ln) of e to the power of x | ln(ex) = x |

e to the power of ln | Eln(x)=x |

Given below is a mathematical expression, evaluate the same using
Log rules.

Log_{2}4+Log_{2}8.

Here, the
numbers 4 and 8 are the exponential numbers while the base is equal
as 2. Here the power rule is applied to the expression followed by
the identity rule. Doing this will provide the final result which
has been explained below:

- Log
_{2}4+Log_{2}8 - =Log
_{2}2^{2}+Log_{2}2^{3} - =2 Log
_{2}2+3Log_{2}2 - =2(1) +3(1)
- =2+3
- Therefore, Log
_{2}8+Log_{2}4=5

Given below is a mathematical expression, solve the same using the
Log rules.

Log_{3}162-Log_{3}2

In this case, the number 162
cannot be expressed as the exponential number that comes with the
base 3. Here, no straight rules are applicable in the direct
manner. However, logarithm rules in this case can actually be
applied in the reverse manner! Take note that these log expressions
can easily be expressed as a single logarithm term via use of
quotient rule in the backward manner.

- Log
_{3}162-Log_{3}2 - =Log
_{3}{162/2} - =Log
_{3}(81) - =Log
_{3}34 - =4 Log
_{3}3 - =4 (1)
- Log
_{3}162- Log_{3}2= 4

Given below is a mathematical expression that needs to be
evaluated using the Log rules. Log_{5}500- 2Log_{5}2+Log_{4}32+Log_{4}8 By the
looks of it, this expression suggests that there are multiple
things occurring at the same time. The first move here is the
simplification of each logarithm number. This will obviously begin
with application of the log rules. When you observe the expression,
there are two different bases involved: 4 and 5.

In this case, you
need to put together the expressions with same base. Now, these
bases with same quotient shall be simplified separately. For the
ones with 5 as base, the Power Rule shall be applied first. After
this, the Quotient Rule shall be applied. For the log with 4 as
base, the Product Rule shall be applied immediately. This will lead
to the final result by addition of the values obtained. So by
applying the process explained above,

- Log
_{5}500-2Log_{5}2+Log_{4}32+Log_{4}8 - =Log
_{5}500-Log_{5}22+Log_{4}432+Log_{4}8 - =Log
_{5}500-Log_{5}4+Log_{4}32+Log_{4}8 - =Log
_{5}(500/4) + Log_{4}(32*8) - =Log
_{5}(125) + Log_{4}(256) - =Log
_{5}(53) + Log_{4}(44) - =3* Log
_{5}5 + 4* Log_{4}(4) - = 3(1) + 4(1)
- =7

Given below is a mathematical expression, solve the same using logarithm rules:

Log

Inside this parenthesis is the product of the factors. Here, you need to apply the Logarithm Product Rule and break them as the sum of separate log expressions. Try to simplify the numerical expressions to derive the exact value when possible. Use the Identity Rule to simplify this process further.

Given below is a mathematical expression, decode the same using
logarithm rules. Log_{7}{49m^{6}/K^{3}} The approach for this mathematical
expression is to apply Logarithm quotient rule. After the
simplification of the expression, the Product Rule needs to be
applied to obtain the sum for logarithm expressions.

- Log
_{7}{49m^{6}/K^{3}} - =Log
_{7}(47m^{6}) – Log_{7}(K^{3}) - = Log
_{7}(49) + Log_{7}(m^{6}) – Log_{7}(K^{3}) - =Log
_{7}(72) +Log_{7}(m^{6}) –Log_{7}(K^{3}) - =2Log
_{7}(72) + 6Log_{7}(m) - 3Log_{7}(k) - =2(1) +6Log
_{7}(m) - 3Log_{7}(k) - =2+ 6Log
_{7}(m) - 3Log_{7}(k)