How to Solve Logarithms

How to Solve Logarithms: a Study guide of its basics.

This article will cover the misunderstood little devil of maths: logarithm. Considered to be tough by most beginners, This little critter is a legend, known for giving students brain cramps. headaches, bad grades and nightmares.Most students are brave and try to take these monsters by the horns.

Generally, in mathematics that is often a good idea, but, if you are new to logarithms, a layman’s approach is a much better option. Here we will learn how to solve logarithms in a simple and fun manner.

Wikipedia defines logarithm as

“The logarithm is the inverse operation to exponentiation. That means the logarithm of a number is the exponent to which another fixed value, the base, must be raised to produce that number.”

Its a well written definition, it defines logarithms in the general manner, so that all bases are covered AND it successfully gives the new student a brain cramp.Hence, the misunderstood devil of maths.

This is where this article comes in handy. This will help the student in tackling the basics of logarithms so that one day, the student can use it in the battle of maths.

Logarithm : the power of powers

consider a number called “x” having a power n such that:

xⁿ = P (base^power = answer)

where P is the result of multiplying x to itself, n number of times.

Now, the logarithm of the above will be:

Log_xP = n

The number whose power was being calculated , will be the base of the logarithm, The value P forms the answer, and n is the power

Log_base(answer) = Power

There you go! So much for the little devil of maths! That above, is the complete mathematical definition of log, in simple english!

Let’s understand this by a few examples!

Consider:

what is the value of Log₃9

Step one, check the base, and the answer.

The base is 3, the answer is 9, so…whats the power????

Its 2 ! there’s your answer!

Log₃9 = 2

Here are a few other examples that the reader can try :

1.log₆₄(4)

2.log₂32

3.log₉81

4.log₃81

5.log₅5

6.log₅1

The first and last one are a little different. in the last part, notice there is “1” in the answer part?? Think of a power that results in the answer in being “1”. Similarly, in the first part, what power of 4 is 64? The answer will be the inverse of that (root).

Section 2: Weapons of logarithm

Logarithms , like other mathematical functions, have a set of properties, which allow us to twist and bend them according to our need. Here are the basic few of those properties.

1. Log_a(mn) = log_a(m) + log_a(n)

If 2 answer values are in multiplication, and they share the same log base, then they can be written as a summation of 2 seperate log values.

Example:

Find the value of log₅(3125)

now we first break down the answer value

3125 = 625*5 or 3125=25*125

lets take the first one and put in the given question

log₅(625*5)

using the 1^st property, we get:

log₅(625) + log₅(5)

now we break down the powers using the 2^nd property.

2.Log_a(m)ⁿ = nlog_a(m) (if answer value has a power, it can be rewritten as power*log_a(answer))

Applying in the question above, we get:

625 = 5⁴, 5 = 5¹

log₅(5)⁴ + log₅(5)¹

4log₅(5) + 1log₅(5)

clearly, we get

4+1 = 5

So, the first to properties were easy enough to understand.

The third one also deals with arithmetic functions within a log.

3. Log_a(m/n) = Log_a(m) – Log_a(n)

When a division occurs within a logarithm, the answer values can be rewritten as seperate log values in subtraction. The value in the denominator will always be the value too be subtracted from the numerator.

Example: Find the value of Log₃(81/27)

Applying 3^rd property

Log₃(81) – Log₃(27)

= 4 – 3 = 1 (since 81 is 3⁴ and 27 is 3³)

4. Log_a(m)ⁿ = nlog_a(m)

A very simple yet incredibly useful rule, can be used to calculate any logarithm as long as m = a. It can also be used to simplify given logarithmic expresseions.

Example:

Find Log₃(81) – Log₃(27) (taken from previous example)

Apply 4^th property.

Log₃(3)⁴ – Log₃(3)³

= 4Log₃(3) – 3Log₃(3)

= 4 – 3 = 1

The answer is the same as in property 3, however using the 4^th property to find the answer is more formal, and is expected to be written as such if such a question is given in an exam.

5. This property comes in handy, when we have to change the base value (a) of the log to the required value (b).

Log_a(m) = Log_b(m)/(Log_b(a))

Notice that, in the denominator, the old base has become the answer value. This allows us to simplify log calculations to a certain degree.This property should be used according to situation.

Example:

Find Log₁₂₅(25)

the base is 125, and the answer is 25. we will not be able to calculate the log by conventional means or deduction by power.

Lets use the 5^th property.

We need a base that will satisfy as a base for the previous base (125) as well as the current answer (25). 5 is the only option.

Log₅(25)/Log₅(125)

2/3 = 0.666

6. Log_a(b) = 1/Log_b(a)

Sometimes, the 5^th property can prove to be too lengthy for a base conversion. Depending on the situation, it can be easier just to switch the base with the answer value.

Example:

Log₆₂₅(5)

using the 5^th property here will be nothing short of stupidity. Lets quicken the pace here –

using 6^th property

we get:

Log₆₂₅(5) = 1/Log₅(625)

= 1/4

= 0.25

Easy, right?

We have covered the basic properties of logarithm, and will be attempting some questions in the next article to gain a deeper understanding of Logarithms.

Who said you can’t tame the dragon??

 

 

 

 

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