# What is Log Base 5 of 125

### 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 \$\$. kapiert.decan do more:

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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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