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Thought-Process to Discover Knowledge

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Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

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mathsExponentsExponents and Logarithm Arithmetics

### Understanding exponents, roots, logarithms

This topic is little advanced for high school students. The various possibilities for roots and logarithms are discussed.

•  root with negative power (root(-2)(4))

•  root with power 0 (root(0)(4))

•  logarithm with fraction base (log_(1/2) 8)

•  logarithm of negative value (log_2 (-4))

•  logarithm with negative base (log_(-2) (-8))

•  logarithm with 0 base (log_0 2)

•  logarithm with 1 base (log_1 2)

Note: Some of the above are not defined and has no meaning. Read through the lesson to understand the details.

click on the content to continue..

revising the basics

Exponent is
(text(base))^(text(power)) = text(exp. result)

If exponentiation result and power are given, then
text(base) = root(text(power))(text(result))
This inverse is called the "root".

If exponentiation-result and base are given, then
text(power) = log_text(base) (text(exp.result))
This inverse is called the "logarithm".

Consider (text(base))^(text(power)) = text(exp. result)

text(base) and text(power) can be one of positive integers, negative integers, 0, 1, or fractions.

Let us consider the important combinations one by one and understand some properties of the inverses "root" and "logarithm".

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(base) and text(power) are positive integers. What is the sign of text(exp. result) ?

• a positive integer
• a positive integer
• can be positive or negative

The answer is "a positive integer".

Consider (text(base))^(text(power)) = text(exp. result)

Given that, text(base) is a negative integer and text(power) is a positive integer. What is the sign of text(exp. result) ?

• a positive integer
• can be positive or negative
• can be positive or negative

The answer is "can be a positive or negative value".

This is explained in the next page.

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(base) is a negative integer and text(power) is a positive integer.

The negative value of the base is shown as -a with a positive integer a.

The exponentiation is
(-a)^b = d
The result d can be a positive or a negative value.

The equation can be simplified to
(-a)^b = (-1)^b a^b = (-1)^b c
and so, the value c is always positive.

In dealing with a negative base, the sign is separated out.

The following are noted.

•  In the inverse "root", the exponentiation result (number c) is a positive number. Negative values for exponentiation result are studied separately in the higher level mathematics (complex numbers).

•  In the inverse "logarithm", the base (number a) is a positive number. Negative bases are not studied, as the negative base is handled as power of -1 as shown.

•  In the inverse "logarithm", the exponentiation result (number c) is a positive number. log of negative values are not defined.

Note 1: If the text(base) is a positive value, then the text(exp. result) is always positive. So logarithm of negative values is not defined.

Note 2: If we encounter log_(text(-ve ) a) (text(+ve ) c) = b or log_(text(-ve ) a) (text(-ve ) c) = b, then convert them into the equivalent exponent equation a^b=c and solve for b.

Given that text(base) is a positive integer and text(power) is a negative integer. What is the sign of text(exp. result) ?

• a negative integer
• a positive fraction
• a positive fraction

The answer is "a positive fraction".

a^(-b) = 1/(a^b).

So, the result is a positive fraction.

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(base) is a positive integer and text(power) is a negative integer.

The negative value of power is shown as -b with a positive integer b.

The exponentiation is
a^(-b) = 1/(a^b) = (1/a)^b = c = 1/d
The result c is a positive fraction.

The following are noted.

•  In the inverse "root", the power is always a positive number. Roots with negative powers are not studied.

Note 1: If root to the negative power is encountered, then convert that into its equivalent exponentiation equation a^b = c and solve for the unknown value.

Note 2: In the inverse "logarithm", the base can be a fraction as given in the equation with 1/a. This is explained along with the generic form of base p/q in the next page.

Given that text(base) is a positive fraction.
a^b = (p/q)^(b) =c

What is the inverse : "logarithm"?

• log_(p/q)(c)
• approach the log as the equivalent exponent
• approach the log as the equivalent exponent

The answer is "approach the log as the equivalent exponent".

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(base) is a fraction.
The exponentiation is
a^b = (p/q)^(b) =c = (p^b)/(q^b)

The result c is a fraction.
The following are noted.

•  The inverse "root", the numerator and denominator can be independently considered as root(b)(c) = root(b)(text(numerator)) / root(b)(text(denominator)).

•  The inverse "logarithm", log_a c, though the base a can be a fraction, it is not formally studied.

If we encounter log_a c where the base a is a fraction, then that is converted into an equivalent equation log_a c = log_b c -: log_b a and solved.

Given that text(power) is a positive fraction.

a^b = a^(p/q) =c

What is the inverse : "root"?

• root(p/q)(c)
• c^(q/p)
• both the above
• both the above

The answer is "both the above".

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(power) is a fraction.
The exponentiation is
a^b = (a)^(p/q) =c = root(q)(a^p)

The following are noted.

•  In the inverse "root", root(b)(c), the power b=p/q can be a fraction. The same is considered as c^(q/p).

•  In the inverse "logarithm", log_a c, the result of this log can be a fraction.

Given that text(base) is 0 and text(power) is a positive-non-zero value.
0^b = 0

What are the inverses "root" and logarithm?

• root(b)(0) = 0
• log to base 0 is not defined
• both the above
• both the above

The answer is "both the above".

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(base) is 0 and text(power) is a positive-non-zero value.
0^b = 0

The following are noted.

•  root(b)(0) = 0.

•  In a^b = 0^b = c = 0, the value of c can never be a non-zero value. So log to the base 0 is not defined for any non-zero values . Further more, log_0 0 = b is also not defined as b can be any number.

Given that text(base) is a positive non-zero value and text(power) is 0.

a^0 = 1

What are the inverses "root" and logarithm?

• root(0)(c) is not defined
• log_a 1 = 0
• both the above
• both the above

The answer is "both the above".

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(base) is a positive non-zero value and text(power) is 0.
a^0 = 1

The following are noted

•  log_a 1 = 0

•  In a^b = a^0 = c = 1, the value of c can never be other than 1. So root(0)(c) is not defined for any value other than 1.

Further more, root(0)(1) = a is also not defined as a can be any number.

Given that text(base) is 1.

1^b = 1

What are the inverses "root" and logarithm?

• root(b)(1) = 1
• log_1 1 is not defined
• both the above
• both the above

The answer is "both the above".

Consider (text(base))^(text(power)) = text(exp. result)

Given that text(base) is 1
1^b = 1

The following are noted.

•  root(b)(1) = 1. some additional learnings will be introduced in the higher mathematics, complex numbers

•  In a^b = 1^b = c = 1, the value of c can never be any value other than 1. So log to the base 1 is not defined for any numbers that is not 1. Further more, log_1 1 = b is also not defined as b can be any number.

Consider (text(base))^(text(power)) = text(exp. result) OR a^b=c

Summary of what we have learned in understanding roots and logarithms.

•  log_(text(-ve ) a) c = b is not formally studied, instead convert to exponent form a^b=c to solve for b.

•  root(text(-ve ) b)(c) = a is not formally studied, instead convert to exponent form a^b=c to solve for a.

•  log_(a=p//q) c = b is not formally studied, instead convert to exponent form a^b = c to solve for b.

•  log_(text(+ve ) a) (text(-ve ) c) = b is not defined, as no value of b satisfies (text(+ve ) a)^b = text(-ve ) c.

•  root(0)(c) = a is not defined, as no value of a satisfies a^0=c for c!=1 and a takes many possible values if c=1

•  log_0 c = b is not defined, as no value of b satisfies 0^b = c for c!=0 and b takes many possible values if c=0.

•  log_1 c = b is not defined, as no value of b satisfies 1^b = c for c!=1 and b takes many possible values if c=1.

slide-show version coming soon