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*