Arithmetics with logarithms, without evaluating the logarithm, is explained. For example `log_a b^m = m log_a b`.

The list of formulas are derived using the first principles of logarithms.

These known results are given as a set of formulas. Students are advised to work them out quickly using the first principles. No need to memorize, and if the formulas are used repeatedly, over time, these can be recalled quickly.

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What is a logarithm?

- logarithm of a number is the power with which the base is raised to get the number
- logarithm is an inverse of exponent
- both the above
- both the above

The answer is "both the above"

What is the value of `log_2 8`?

- by first principles of logarithm `log_2 8 = log_2 (2^3) = 3`
- by first principles of logarithm `log_2 8 = log_2 (2^3) = 3`
- `2xx8 = 16`

The answer is "`log_2 8 = log_2 (2^3) = 3`".

By first principles, the value is expressed as an exponent of base

`log_2 8 = log_2 2^3`

The exponent, to which base is raised to get a number, is the logarithm to the base of the number.

`log_2 8 = log_2 2^3 = 3`

`2` is the base

`8` is the number

`3` is the log of the number

What is the exponent to which `2` is raised to get value `8`?

- `log_2 8`
- `log_2 8`
- `2xx8`

The answer is "`log_2 8`"

We learned the first principles of logarithm as, for any value `x` and base `b`,

`log_b x = n hArr b^n = x`

In this topic, the above is used to understand some known results. *These known results are given as a set of formulas. Students are advised to work them out quickly using the first principles. No need to memorize, and if the formulas are used repeatedly, over time, these can be recalled quickly.*

It is noted that `2^4 = 16` and `(-2)^4 = 16`, so `4th` root of `16` has more than one possible values.

That is generalized as `p^n = q^n` does not imply that `p=q`.

If the base `2` is the same, then to get the value `16`, the exponent is unique `4`. That is only `2^4 = 16` for base `2` and number `16`.

If `log_b x = log_b y` then is it true that `x = y`?

Note that by first principles,

`color(coral)(log_b x = n hArr b^n=x)`

`color(deepskyblue)(log_b y = n hArr b^n=y)`

So,

`color(coral)(b^n) `` = color(deepskyblue)(b^n)`

`color(coral)(x) `` = color(deepskyblue)(y)`

- No, `x` and `y` need not be equal
- yes,`x=y`
- yes,`x=y`

The answer is "if base of the logarithm is same, then `x=y`".

Generalizing this, for real numbers `x`, `y` and `b`, *`log_b x = log_b y hArr x = y`*

What is `log_2 2^10`?

Note that the value is given as an exponent of base.

- `10`
- `10`
- `1024`

The answer is "`10`".

Generalizing this, for real numbers `b` and `n`, *`log_b b^n = n`*

What is `2^(log_2 8)`?

Note that by first principles,

`color(coral)(log_2 8 = 3)`

`color(deepskyblue)(2^3 = 8)`

So,

`color(deepskyblue)(2)^(color(coral)(log_2 8)) = 2^3`

The question is, what is the simplified form of the above expression?

- `2^8`
- `8`
- `8`

The answer is "`8`".

Generalizing this, for real numbers `x` and `b`, *`b^(log_b x) = x`*

What is `log_2 1`?

Note that `2^0 = 1`

- `1`
- `0`
- `0`

The answer is "`0`".

Generalizing this, for a real number `b`, *`log_b 1 = 0`*

What is `log_2 2`?

Note that `2^1 = 2`

- `1`
- `1`
- `0`

The answer is "`1`".

Generalizing this, for a real number `b`, *`log_b b = 1`*

It is given that `log_2 32 = 5` and `log_2 16 = 4`.

What is `log_2 (32 xx 16)`?

Note that by first principles,

`color(coral)(log_2 32 = 5 hArr 2^5=32)`

`color(deepskyblue)(log_2 16 = 4 hArr 2^4 =16)`

So,

`log_2 (color(coral)(32) ``xx color(deepskyblue)(16) )`

`= log_2 (color(coral)(2^5) ``xx color(deepskyblue)(2^4) )`* substituting the known result `2^5 xx 2^4 = 2^(5+4)`*

`= log_2 2^(color(coral)(5) + color(deepskyblue)(4) )`

- `5+4`
- `5+4`
- `5 xx 4`

The answer is "`5+4`".

That is `log_2 (color(coral)(32) ``xx color(deepskyblue)(16) )` `= log_2 (color(coral)(32) )``+ log_2 (color(deepskyblue)(16) )`

Generalizing this, for real numbers `x`, `y` and `b`, *`log_b(xy) = log_b x + log_b y`*

It is given that `log_2 32 = 5` and `log_2 16 = 4`.

What is `log_2 (32 -: 16)`?

Note that by first principles,

`color(coral)(log_2 32 = 5 hArr 2^5=32)`

`color(deepskyblue)(log_2 16 = 4 hArr 2^4 =16)`

So,

`log_2 (color(coral)(32) ``-: color(deepskyblue)(16) )`

`log_2 (color(coral)(2^5) ``-: color(deepskyblue)(2^4) )`* substituting the known result `2^5 -: 2^4 = 2^(5-4)`*

`log_2 2^(color(coral)(5) - color(deepskyblue)(4) )`

The question is, what is the simplified form of the above expression?

- `5-4`
- `5-4`
- `5 -: 4`

The answer is "`5-4`".

That is `log_2 (color(coral)(32) ``-: color(deepskyblue)(16) )` `= log_2 (color(coral)(32) )``- log_2 (color(deepskyblue)(16) )`

Generalizing this, for real numbers `x`, `y` and `b`, *`log_b(x/y) = log_b x − log_b y`*

What is `log_2 (8^5)`?

Note that by first principles,

`color(coral)(log_2 8 = 3 hArr 2^3=8)`

`log_2 color(coral)(8)^5 `

`= log_2 (color(coral)(2^3))^5) `

`= log_2 2^(color(coral)(3)xx5)`

`= color(coral)(3) xx 5`

`= 5 xx color(coral)(3)`*substituting that `log_2 8 = 3`*

`=5 xx color(coral)(log_2 8)`

- `5 log_2 8`
- `5 log_2 8`
- `log_2 (5xx8)`

The answer is "`5 log_2 8`".

Generalizing this, for real numbers `x`,`n` and `b`, *`log_b x^n = n log_b x`*

It is noted that

`b^n = x hArr log_b x = n`

`color(coral)(2^3 = 8) hArr root(3)(8) = color(deepskyblue)(8^(1/3)= 2)`

There are two exponents in the above `color(coral)(2^3)` and `color(deepskyblue)(8^(1/3))`

applying logarithm for the exponent `color(coral)(2^3 = 8)` with `b=2` and `x=8`, we get

`color(coral)(log_2 8 = 3)`

applying logarithm for the exponent `color(deepskyblue)(8^(1/3)= 2)` with `b=8` and `x =2`

`color(deepskyblue)(log_8 2 = 1/3)`

Comparing the two logarithms, which of the following is correct?

- `log_2 8 = log_8 2`
- `log_2 8 = 1 // log_8 2`
- `log_2 8 = 1 // log_8 2`

The answer is "`log_2 8 = 1 // log_8 2`".

Generalizing this, for real numbers `a` and `b`, *`log_b a = 1 -: (log_a b)`*

It is noted that

`color(coral)(2^12) = color(deepskyblue)(4^6)`

Converting the left hand side `color(coral)(2^12)` to log form

`color(coral)(log_2 (2^12) = 12)`

The same in right hand side

`color(coral)(log_2 (4^6)) = color(deepskyblue)(6) xx color(coral)(log_2(4))`

Converting the right hand side `color(deepskyblue)(4^6)` to log form

`color(deepskyblue)(log_4 (4^6) = 6)`

substituting the value `6`

`color(coral)(log_2 (4^6)) = color(deepskyblue)(log_4 (4^6)) xx color(coral)(log_2(4))`

What does the above prove?

- `log_2(2^12) = log_4(2^12) xx log_2 4`
- `log_2(2^12) = log_4(2^12) -: log_4 2`
- both the above
- both the above

The answer is "both the above".

Generalizing this, for real numbers `x`, `a` and `b`, *`log_b x = log_a x xx log_b a`**`log_b x = log_a x -: log_a b`*

**Known results in Logarithms** :

`log_b x = n hArr b^n = x`

`log_b x = log_b y hArr x = y`

`log_b b^n = n`

`b^(log_b x) = x`

`log_b 1 = 0`

`log_b b = 1`

`log_b(xy) = log_b x + log_b y`

`log_b(x/y) = log_b x − log_b y`

`log_b x^n = n log_b x`

`log_b x = (log_a x) -: (log_a b)`

`log_b x = log_b a xx log_a x`

*Solved Exercise Problem: *

What is `log_3 (3^(-2))`?

- `1/9`
- `-2`
- `-2`

The answer is "`-2`"

*slide-show version coming soon*