Converting Exponential Form to Logarithmic Form

Exponential equations and logarithmic equations describe the exact same relationship between three numbers, just written in a different order. Once you see the pattern, converting from exponential form to logarithmic form becomes a quick, mechanical step rather than something to memorize by rote.

An exponential equation has a base raised to a power, set equal to a result:

\( b^{x} = y \)

Here \(b\) is the base, \(x\) is the exponent, and \(y\) is the result. If you have not seen this relationship introduced yet, it helps to first review what a logarithm actually is before working through conversions.

The logarithmic form of that same relationship is:

\( \log_{b}(y) = x \)

This says: "the logarithm, base \(b\), of \(y\) equals \(x\)." It is asking the exact same question as the exponential equation, just phrased differently: "what power do I raise \(b\) to, in order to get \(y\)?"

To rewrite \( b^{x} = y \) in logarithmic form, follow three moves:

1. The base \(b\) stays the base, but it now sits as a subscript on \(\log\).
2. The result \(y\) (the number on the right side of the exponential equation) moves inside the logarithm, as the argument.
3. The exponent \(x\) moves to the other side of the equals sign, by itself.

So the general conversion is:

\( b^{x} = y \quad \Longleftrightarrow \quad \log_{b}(y) = x \)

Notice that the base never changes position relative to the other two numbers, it is always attached to the log. The exponent and the result simply swap sides of the equation.

Example 1: Convert \( 2^{3} = 8 \) to logarithmic form.

Here \(b = 2\), \(x = 3\), and \(y = 8\). Applying the rule:

\( \log_{2}(8) = 3 \)

Example 2: Convert \( 5^{2} = 25 \) to logarithmic form.

Here \(b = 5\), \(x = 2\), and \(y = 25\), so:

\( \log_{5}(25) = 2 \)

Example 3: Convert \( 4^{-2} = \frac{1}{16} \) to logarithmic form.

The exponent does not need to be positive for the rule to work. With \(b = 4\), \(x = -2\), and \(y = \frac{1}{16}\):

\( \log_{4}\left(\frac{1}{16}\right) = -2 \)

Example 4: Convert \( x^{4} = 81 \) to logarithmic form, where \(x\) is the unknown base.

This time the letter \(x\) is being used as the base, not the exponent, so be careful to match the roles rather than the labels: base is \(x\), exponent is \(4\), result is \(81\):

\( \log_{x}(81) = 4 \)

Two bases show up so often that they get their own shorthand notation.

When the base is 10, the equation \( 10^{x} = y \) converts to \( \log(y) = x \), where \(\log\) with no written base means base 10. This is called the common logarithm.

When the base is \(e\) (Euler's number, approximately 2.718), the equation \( e^{x} = y \) converts to \( \ln(y) = x \). This is the natural logarithm, and \(\ln\) always implies base \(e\) without writing it.

For example, \( 10^{2} = 100 \) becomes \( \log(100) = 2 \), and \( e^{1} = e \) becomes \( \ln(e) = 1 \).

Because \( 2^{x} = y \) and \( \log_{2}(y) = x \) describe the same numbers, the graph of \( y = \log_{2}(x) \) climbs slowly and passes through \((1, 0)\), since \( \log_{2}(1) = 0 \) matches \( 2^{0} = 1 \).

Watch out for these slip-ups when converting:

• Putting the exponent inside the log instead of the result. Remember, it is always the result \(y\) that becomes the argument of the logarithm, and the exponent \(x\) that ends up alone on the other side.
• Losing track of the base when it is a variable or a fraction. The base of the exponential expression is always the base (subscript) of the logarithm, no matter what it looks like.
• Forgetting that a negative exponent is still valid. It simply produces a negative value on the logarithmic side of the equation.

Try converting these on your own before checking: \( 3^{4} = 81 \), \( 7^{0} = 1 \), and \( 6^{-1} = \frac{1}{6} \). In each case, identify the base, the exponent, and the result first, then slot them into \( \log_{b}(y) = x \).

Once this conversion feels comfortable, it is worth practicing it in reverse by learning how to go from logarithmic form back to exponential form, since many problems ask you to move fluidly in both directions. From there, this skill becomes the foundation for solving logarithmic equations, where rewriting a log equation as an exponential one is often the very first step toward isolating the variable.

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