\[\begin{align*} \displaystyle\boxed{\log_a b = x \text{ is equivalent to } a^x = b} \end{align*}\]
What is $\log_a b$?
This expression reads “log base a of b”. In mathematics, $\log_a b$ is called the logarithm of $b$ with base $a$. The logarithm $\log_a b$ answers the question:
“What power do we need to raise $a$ to, in order to get $b$?”
\[\begin{align*} \displaystyle\log_a b = x \quad \text{means that} \quad a^x = b \end{align*}\]
In other words, $\log_a b = x$ means that when we raise $a$ to the power $x$, we get $b$:
\[\begin{align*} \displaystyle a^x = b \end{align*}\]
Example. Compute $\log_5 125$.
- We set $\log_5 125 = x$.
- This means we are looking for $x$ such that $5^x = 125$.
- We know $5^3 = 125$, so $x = 3$.
Thus,
\[\begin{align*} \log_5 125 = 3 \end{align*}\]
Key Points to Remember:
- Base $a$: The number we are raising to a power (in our example, $a = 5$).
- Argument$b$: The number we want to get by raising $a$ to some power (in our example, $b = 125$).
- $\log_a b = x$ is the exponent, such that $x$ that satisfies $a^x = b$.
\[\begin{align*} \displaystyle\boxed{\log_a b = x \text{ is equivalent to } a^x = b} \end{align*}\]
Similarly,
\[\begin{align*}\displaystyle\log_{0.5} 2 = -1 \quad \text{because} \quad (0.5)^{-1} = 2 \\ \displaystyle\log_{2} 32 = 5 \quad \text{because} \quad 2^5 = 32 \\ \displaystyle\log_{3} 81 = 4 \quad \text{because} \quad 3^4 = 81\\ \displaystyle\log_{12} 1 = 0 \quad \text{because} \quad 12^0 = 1 \\ \end{align*}\]
