What's the Relationship Between A^x = B and Log_a B = X?

The equations $a^x=b$ and $\log_a b=x$ are equivalent: they state the same relationship in exponential and logarithmic forms. A logarithm gives the exponent to which a base must be raised to produce a specified value.

The Reasoning / Mechanism

In exponential form,

$$a^x=b,$$

$a$ is the base, $x$ is the exponent, and $b$ is the result. Rewriting this in logarithmic form gives

$$\log_a b=x.$$

This is not a new equation. The expression $\log_a b$ means “the exponent of $a$ that produces $b$.” The roles remain fixed during conversion:

For example,

$$2^3=8 \quad \Longleftrightarrow \quad \log_2 8=3.$$

Both statements say that raising $2$ to the power $3$ gives $8$. To convert either way, identify the base, exponent, and result, then change the notation without changing their values.

For real logarithms, the domain conditions are

$$a>0, \qquad a\neq 1, \qquad b>0.$$

The base must be positive and cannot equal $1$. The argument must be positive because $a^x>0$ for every real $x$ when $a>0$. For example, has no real value.

Exponential and logarithmic functions are inverse functions:

$$\log_a\left(a^x\right)=x, \qquad a^{\log_a b}=b.$$

To solve $5^x=17$, rewrite it as $x=\log_5 17$. Using change of base,

$$x=\frac{\ln 17}{\ln 5}\approx 1.76.$$

Substitution checks the result: $5^{1.76}$ is approximately $17$. A common misconception is that $\log_a b$ means $a$ divided by . It does not: a logarithm represents an exponent.

IB Exam Technique

This is SL 1.5: Introduction to logarithms, studied in AA SL and HL. On Paper 1, show the conversion and retain exact forms where possible. On Paper 2, technology may evaluate logarithms, but you should still show the equation and change-of-base setup. Always reject answers that violate the logarithm's domain conditions.