Why is 1 log 1 not 1

Looking back: potentiating and pulling roots

The calculation of a power of the type $$ b ^ x $$ is called exponentiation.

Example: $$ 2 ^ 4 = 2 * 2 * 2 * 2 = 16 $$

But what if you are looking for the base? You can reverse the exponentiation.

Example: $$ root 4 16 = 2 $$, because $$ 2 * 2 * 2 * 2 = 16 $$

The extraction of roots is the reverse of exponentiation.

What is the logarithm?

What if you want to find the exponent?

Example:

$$ 2 ^ x = 16 $$

How do you have to raise $$ 2 $$ to the power to get to $$ 16 $$? You already know the result: it's $$ 4 $$.

That is exactly what the logarithm does.

The logarithm of $$ 16 $$ to base 2 is the number with which you raise $$ 2 $$ to the power to get $$ 16 $$.

Write: $$ log_2 16 = 4 $$, because $$ 2 ^ 4 = 2 * 2 * 2 * 2 = 16 $$.

Read: The logarithm of $$ 16 $$ to the base $$ 2 $$ is equal to $$ 4 $$.

With the logarithm you determine the exponent.

Definition of logarithms

Let $$ y $$ and $$ b ≠ 1 $$ two positive numbers. Then the logarithm of $$ y $$ to the base $$ b $$ is the number $$ x $$ with which one has to raise $$ b $$ to the power to get $$ y $$.

To solve the equation $$ 2 ^ x = 16 $$, write $$ log_2 16 $$.
The expressions $$ 2 ^ x = 16 $$ and $$ log_2 16 $$ are therefore synonymous.

So:
$$ log_2 16 = 4 $$, since $$ 2 ^ 4 = 2 * 2 * 2 * 2 = 16 $$.

Pocket calculators can calculate $$ log_2 16 $$. Try it. It will display $$ 4 $$.

$$ b ^ x = y $$ means the same as $$ log_b y = x $$.

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Examples (1)

You can solve these examples without a calculator just by thinking:


$$ log_10 10000 = 4 $$, since $$ 10 ^ 4 = 10 * 10 * 10 * 10 = 10000 $$


$$ log_10 10/1000 = -3 $$, since $$ 10 ^ -3 = 1 / [10 ^ 3) = 1/1000 $$


$$ log_3 81 = 4 $$, since $$ 3 ^ 4 = 3 * 3 * 3 * 3 = 81 $$


$$ log_ {1/4} 2 = -1 / 2 $$,
since $$ (1/4) ^ (- 1/2) = (1 / (4 ^ -1)) ^ (1/2) = 4 ^ (1/2) = sqrt (4) = 2 $$

$$ b ^ x = y $$ means the same as $$ log_b y = x $$.

Examples (2)

You can solve the following examples without a calculator just by thinking:

$$ log_5 5/25 = -2 $$,
da $$ 5 ^ x = 1/25 $$ ⇔ $$ 5 ^ x = 1/5 ^ 2 $$ ⇔ $$ 5 ^ x = 5 ^ -2 $$ ⇔ $$ x = -2 $$


$$ log_5 125 = 3 $$,
because $$ b ^ 3 = 125 $$ ⇔ $$ b ^ 3 = 5 ^ 3 $$ ⇔ $$ b = 5 $$


$$ log_3 343 = 7 $$,
because $$ 7 ^ 3 = y $$ ⇔ $$ y = 343 $$


$$ log_9 1 = 0 $$,
because $$ 9 ^ 0 = y $$ ⇔ $$ y = 1 $$


$$ log_3 sqrt3 = 0.5 $$,
since $$ 3 ^ x = sqrt (3) $$ ⇔ $$ 3 ^ x = 3 ^ 0.5 $$ ⇔ $$ x = 0.5 $$

$$ b ^ x = y $$ means the same as $$ log_b y = x $$.

Power laws:

$$ 1 / a ^ n = a ^ -n $$

$$ sqrta = a ^ (1/2) = a ^ 0.5 $$

What else is there to know?

$$ a) $$ Logarithms of negative numbers do not exist because $$ b ^ x $$ is always positive if $$ b> 0 $$. $$ y $$ can therefore not take the value 0.

$$ b) $$ Since the base 10 logarithm is often used, one also writes $$ log_10 (y) = log (y) $$ as a convention.

You can leave out the 10 as a base.

$$ c) $$ $$ b ^ 1 = b $$, i.e. $$ log_b b = 1 $$ for all $$ b> 0 $$.

$$ d) $$ $$ b ^ 0 = 1 $$, i.e. $$ log_b 1 = 0 $$ for all $$ b> 0 $$.

$$ d) $$ The case $$ b = 1 $$ is excluded because $$ 1 ^ x $$ is always the same as $$ 1 $$ for all $$ x> 0 $$.

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