Why Logarithms: The Inverse of Exponentials

Logarithms as “Solving for the Exponent”

Exponential equations often hide the unknown in the exponent. For example, if you know 2^3 = 8, you can answer “what power of 2 gives 8?” by inspection. But if you face something like 2^x = 20, the unknown is still the exponent, and ordinary algebraic moves (like dividing both sides by 2) do not isolate x. Logarithms are designed specifically to answer: what exponent produces a given value?

(1) Definition: log_b(x) is the exponent

The logarithm base b of x is defined as the exponent y you must raise b to in order to get x:

log_b(x) = y if and only if b^y = x

This is not a rule to memorize; it is a translation statement. It tells you that logarithms and exponentials are two ways to say the same relationship.

(2) Translating between exponential and logarithmic forms

To translate fluently, match each part of the definition:

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• b is the base
• x is the result (the output of the exponential)
• y is the exponent

Template:

• Log form: log_b(x) = y
• Exponential form: b^y = x

Step-by-step translation (log → exponential):

1. Identify the base b (the subscript).
2. Identify the log’s input x (inside parentheses).
3. Set up b^(answer) = x.

Example A: Convert log_5(125) = 3 to exponential form.

• Base is 5, input is 125, output is 3
• So: 5^3 = 125

Step-by-step translation (exponential → log):

1. Identify the base b.
2. Identify the exponent y.
3. Identify the result x.
4. Write log_b(x) = y.

Example B: Convert 10^{-2} = 0.01 to log form.

• Base is 10, exponent is −2, result is 0.01
• So: log_10(0.01) = -2

(3) Domain restrictions and valid bases

For real-valued logarithms, there are two key restrictions built into the definition.

Restriction on the input: x > 0

• log_b(x) asks for a real exponent y such that b^y = x.
• When b > 0, the exponential expression b^y is always positive for real y.
• So it can never equal 0 or a negative number, meaning log_b(0) and log_b(-3) are not real numbers.

Restrictions on the base: b > 0 and b ≠ 1

• If b ≤ 0, expressions like b^y are not real for many real exponents y, so the logarithm would not be a well-defined real function.
• If b = 1, then 1^y = 1 for every real y, so you could never solve 1^y = x for any x ≠ 1. The “inverse” would fail.

Quick validity check:

(4) Inverse relationship and the graph idea (reflection across y = x)

The definition log_b(x) = y ⇔ b^y = x says that logarithms “undo” exponentials with the same base. In function language:

• f(x) = b^x
• f^{-1}(x) = log_b(x)

Conceptually, inverse functions swap inputs and outputs. That swapping has a clear geometric meaning: the graph of a function and the graph of its inverse are reflections of each other across the line y = x.

How to see the reflection idea without heavy graphing:

• If b^2 = 9, then the point (2, 9) lies on the graph of y = b^x.
• The inverse relationship says log_b(9) = 2, so the point (9, 2) lies on the graph of y = log_b(x).
• The points (2, 9) and (9, 2) are mirror images across y = x because their coordinates are swapped.

This “swap” also matches the domain restriction: exponentials output only positive values, so logarithms can only accept positive inputs.

(5) Evaluating simple logarithms by recognizing powers

Many logarithms can be evaluated exactly by rewriting the input as a power of the base. The guiding question is always:

“What exponent on b gives x?”

Step-by-step method (recognize a power):

1. Write the problem: log_b(x).
2. Rewrite x as b^(something) if possible.
3. The “something” is the value of the log.

Example 1: log_2(8)

• Recognize: 8 = 2^3
• So: log_2(8) = 3

Example 2: log_10(0.01)

• Recognize: 0.01 = 10^{-2}
• So: log_10(0.01) = -2

Example 3: log_3(1/27)

• Recognize: 1/27 = 3^{-3} because 3^3 = 27
• So: log_3(1/27) = -3

Example 4: log_4(2)

• Recognize: 4 = 2^2, so ask: what power of 4 gives 2?
• Write 2 as a power of 4: 2 = 4^{1/2} because 4^{1/2} = sqrt(4) = 2
• So: log_4(2) = 1/2

Practice set (evaluate by recognition):

• log_5(25)
• log_2(1/16)
• log_{10}(1000)
• log_3(81)
• log_7(1)

Self-check answers:

• log_5(25) = 2 because 5^2 = 25
• log_2(1/16) = -4 because 2^{-4} = 1/16
• log_{10}(1000) = 3 because 10^3 = 1000
• log_3(81) = 4 because 3^4 = 81
• log_7(1) = 0 because 7^0 = 1