Logarithms as inverse operations

A logarithm names an exponent, including an unfamiliar one

The preceding lesson established that the standard real exponential bᵗ, with b > 0 and b ≠ 1, is continuous and strictly monotone on all real t, with range all positive real values. For every positive x, there is therefore one and only one real t such that bᵗ = x. The logarithm gives that exponent a name: log_b(x) = t means exactly bᵗ = x. The base b, the positive argument x and the resulting exponent t have different roles.

For example, log₂(16) = 4 because 2⁴ = 16, while log₂(1/8) = −3 because 2⁻³ = 1/8. An exponent need not be an integer or a familiar fraction. The exact expression log₂(5) denotes the unique real t with 2ᵗ = 5. Since 2² < 5 < 2³ and 2ᵗ is increasing, 2 < log₂(5) < 3. Leaving this value as a logarithm records an exact number rather than an incomplete or missing answer.

We are inverting the exponential function, not dividing by its base or counting only whole multiplications. Fractional and negative powers are already part of the standard exponential, and irrational exponents are part of its continuous real extension. Existence and uniqueness cover every positive argument, beyond a finite table of powers. A decimal estimate can be useful later, but it is not the definition and does not replace an exact identity.

The base and argument have separate conditions

The standard real logarithm used here requires b > 0, b ≠ 1 and x > 0. A base of one gives 1ᵗ = 1 at every real t: argument 1 would have many exponents and other arguments none, so it has no single-valued logarithmic inverse. A zero or negative base does not supply a positive-valued exponential defined for every real exponent. These exclusions concern the base, even when a particular integer power can be calculated.

An argument of zero or a negative argument has no real logarithm in this lesson, because a positive-base real exponential never gives those values. A negative logarithmic output is different: log₂(1/8) = −3 is valid, with positive argument 1/8. Nor must the surrounding input variable itself be positive. In log₂(x + 4), it is x + 4 that must be positive, so x = −3 is permitted and gives logarithm 0.

Keep the argument grouped. The expressions log₂(x²), (log₂ x)² and 2 log₂ x have different original domains and meanings. The first logs the square and admits x ≠ 0; the latter two first log x and require x > 0. Parentheses also distinguish log₂(2x) from (log₂ 2)x. We will rewrite expressions only after checking every inner value and original argument. Complex logarithms require a different framework and are outside the real functions considered here.

Recognise a power before reaching for an approximation

The defining exponential identity gives log_b(1) = 0, log_b(b) = 1 and log_b(bᵘ) = u for every real u and every valid base. In particular, log₈(2) = 1/3 because 8^(1/3) = 2, and log₉(1/3) = −1/2 because 9^(−1/2) = 1/3. A root or reciprocal in the positive argument can correspond to a fractional or negative exponent without making the logarithm invalid.

A base below one works in the same definition. Since (1/3)² = 1/9, log_(1/3)(1/9) = 2. Since (1/3)⁻² = 9, log_(1/3)(9) = −2. These signs reverse the familiar base-above-one pattern. Writing the exponential statement is a reliable check: retain the stated base and ask which exponent produces the stated positive argument.

Some exact values do not simplify to familiar fractions. For instance, log₂(3) is the unique exponent producing 3; 1 < log₂(3) < 2 follows from 2 < 3 < 4. In fact it is irrational: if it were p/q in lowest terms with positive integers p and q, then 2ᵖ = 3ᑫ, contradicting unique prime factorisation. This particular proof does not claim that every logarithm of an unfamiliar argument is irrational. Exact named logarithms, known rational exponents and later numerical approximations have distinct uses.

Positive arguments give every real output

For f(x) = log_b x on its full domain x > 0, the complete range is all real y. Given any proposed real y, choose x = bʸ. This input is positive and log_b(bʸ) = y, so every claimed output is attained. Conversely, the definition supplies a real exponent for every permitted positive input. The sole preimage of y is bʸ, by the one-to-one property of the exponential.

If b > 1, the exponential is strictly increasing and its inverse logarithm is strictly increasing as well: x₁ < x₂ implies log_b x₁ < log_b x₂. Thus positive arguments below 1 give negative logarithms, argument 1 gives zero and arguments above 1 give positive logarithms. If 0 < b < 1, the inverse is strictly decreasing: the same input inequality gives the opposite output inequality, and the signs on either side of argument 1 reverse.

For b > 1, log_b x decreases without bound as x approaches zero from the positive side and increases without bound as x grows without bound. For 0 < b < 1 the directions reverse. The line x = 0 is a vertical asymptote of the full parent logarithm, rather than an input at which a real infinity is assigned. These statements follow by reversing the exponential's end behaviour. An additional restriction such as x ≥ 1 changes the function's actual range and which boundary behaviour its own domain can approach.

Inverse identities keep the inner function's domain

There are two inverse identities with different input domains: log_b(bᵗ) = t for every real t, and b^(log_b x) = x for x > 0. In the first, the inner exponential is always defined and positive. In the second, the inner logarithm must first be defined; its cancellation cannot create an original value at x = 0 or x < 0. An inverse operation returns the starting value when the full composition is permitted.

For 2^(log₂(x − 3)), the original domain is x > 3, and the value is x − 3 there. The simpler rule x − 3 on all real inputs would be an extension. Similarly, log₂(2^(x²)) = x² has all real inputs because its inner exponential is always positive. The domain belongs to the complete original expression, not only its final appearance after cancellation.

Restrictions on an exponential also restrict its inverse. The function E(t) = 2ᵗ on 0 ≤ t ≤ 4 has range [1, 16]; its inverse is log₂ x on 1 ≤ x ≤ 16, with range [0, 4]. The parent logarithm at x = 32 still has mathematical value 5, but input 32 is outside this restricted inverse's domain and its formal output 5 is outside the restricted inverse's range. The inverse notation f⁻¹ means an inverse function here, not the reciprocal 1/f.

An inner rule can change the number of preimages

For F(x) = log₂(x − 3), the argument condition gives x > 3. Every real y is attained at x = 3 + 2ʸ, which is permitted, so F has range all real y with one preimage for each. For G(x) = 1 + log₂ x restricted to 1/2 ≤ x ≤ 4, strict increase gives endpoint outputs 0 and 3. Conversely, any y in [0, 3] has the permitted preimage x = 2^(y − 1), proving the whole range [0, 3]. A displayed selection of inputs alone cannot prove that intervening outputs are attained.

For H(x) = log₂(x²), the condition x² > 0 gives x ≠ 0. The inner square allows positive and negative inputs, and H is not one-to-one on this whole domain. For any real y, x² = 2ʸ gives exactly two permitted preimages x = ±2^(y/2). Neither is zero, and both give H(x) = y. Thus its range is all real y even though it has two inverse branches; restricting to x > 0 would select the positive branch.

For J(x) = log₂√x, the square root first requires x ≥ 0 and the logarithm then requires √x > 0, leaving x > 0. Its sole preimage of any real y is x = 2^(2y), proving an all-real range. For K(x) = log₂((x − 1)/(x − 1)), the inner quotient first excludes x = 1. At every other real x it equals 1, so K has range {0}, with the entire original domain as the preimage of zero. Cancelling the inner quotient never restores x = 1.

Products and quotients turn into sums and differences on a shared domain

Let b be a valid base and let U > 0 and V > 0. If p = log_b U and q = log_b V, then U = bᵖ and V = bᑫ. The exponential law gives UV = b^(p + q), so the unique logarithm of UV is p + q. Similarly, U/V = b^(p − q), giving log_b(UV) = log_b U + log_b V and log_b(U/V) = log_b U − log_b V. The two separate logarithms require both separate arguments to be positive.

A positive product or quotient alone does not guarantee that separate logarithms exist. For example, log₂((−2)(−4)) = log₂ 8 = 3 is defined, while log₂(−2) + log₂(−4) is not a real expression. Likewise, log₂((−2)/(−1)) = 1 is defined, but the corresponding difference of the two real logarithms is undefined. Expanding without the separate positivity conditions can discard an entire part of the original domain.

When condensing a sum of defined logarithms, retain the sum's original domain even if the single logarithm of the product admits other inputs. There is no parallel addition law inside an argument: log₂(1 + 1) = 1, whereas log₂ 1 + log₂ 1 = 0. A product law concerns multiplication of arguments, not addition of them. Subtraction inside an argument likewise does not license subtracting their logarithms.

A power law needs a positive base for the separate logarithm

For U > 0, any real r and a valid logarithm base b, the standard positive-base power gives log_b(Uʳ) = r log_b U. To see this, put p = log_b U, so U = bᵖ. The positive-base exponential identity Uʳ = (bᵖ)ʳ = b^(pr) gives logarithm pr. This covers positive, negative, zero and irrational r while keeping U strictly positive; Uʳ also stays positive, so the outer logarithm is defined.

An integer or rational power of a negative input can be defined under the preceding lesson's real-root conventions, but that does not make log_b of that negative input defined. For x ≠ 0, the correct whole-domain identity is log_b(x²) = 2 log_b|x|, since |x| > 0 and x² = |x|². The expression 2 log_b x instead applies only on x > 0. At x = −2, log₂(x²) = 2 while 2 log₂ x is undefined. An absolute value here records the required positive magnitude; it does not assign a logarithm at zero.

For the composed root, log_b√x = (1/2) log_b x on x > 0. Although √0 exists, its logarithm does not. A reciprocal power such as log_b(U⁻¹) = −log_b U also retains U > 0. Preserve prior denominator, root and context restrictions in more complicated expressions. A formula that extends a simplification to additional inputs describes a different function; a correct equality on a shared domain does not erase that domain.

Changing the base changes the unit of exponent

Let b > 0, b ≠ 1, c > 0, c ≠ 1 and x > 0. Put y = log_b x, so x = bʸ. Applying the positive-base power law in base c gives log_c x = y log_c b. The denominator log_c b is not zero: a logarithm in a valid base equals zero only at argument 1, and b ≠ 1. Therefore log_b x = (log_c x)/(log_c b). The numerator records the argument; the denominator records the original base.

For example, log₄ 8 = (log₂ 8)/(log₂ 4) = 3/2. For a base below one, log_(1/4) 8 = 3/(−2) = −3/2. Setting b = 1/2 and c = 2 gives log_(1/2) x = −log₂ x on x > 0. The minus sign explains why decreasing and increasing parent logarithms with reciprocal bases have opposite values; it does not change their common positive-argument domain.

Common logarithm means base 10 and is often written log x; an unspecified log can have a different convention in another subject or software tool, so this lesson states its bases. Natural logarithm is written ln x and means log_e x, where e is the standard constant greater than one used in natural exponentials. Its construction and special role in continuous rates and calculus come later. The exact identities log_b x = ln x/ln b and log_b x = log₁₀ x/log₁₀ b use the same guards. A calculator's decimal output approximates the exact ratio and must not decide exact domain or equality by rounding.

A logarithmic ratio scale converts a factor into an added step

A mathematical scale can be stipulated as S(A) = log_b(A/Aref), with A > 0 and fixed Aref > 0 measured in the same units. The ratio A/Aref is dimensionless; the logarithm names the exponent producing that ratio. The reference gives S(Aref) = 0. This scale need not claim a new physical law: it is a defined way of representing positive amounts relative to a specified reference.

For two permitted amounts A₁ and A₂, the quotient law gives S(A₂) − S(A₁) = log_b(A₂/A₁). Multiplying an amount by positive factor k adds log_b k to the scale when both original amounts are permitted. In base 2, doubling adds one and halving subtracts one. An increase of 20% instead adds the exact value log₂(6/5), not 0.2; the percentage is a factor change of the amount, not an additive scale step.

Changing the reference or measurement units requires care. Converting both A and Aref from litres to millilitres leaves the ratio unchanged and hence leaves S unchanged. Changing only the reference changes the scale's zero and gives an additive shift. Keep restrictions on permitted amounts when finding the actual range or a preimage. Established scientific logarithmic scales can include their own coefficients and conventions; the fictional scale defined here does not supply claims about earthquake energy, hearing, chemistry or measured mechanisms.

Check the selected input and proposed output independently

The explorer uses the seven original rules in the domain-and-preimage table. Select an input and a proposed output separately. The input check first evaluates the inner expression and checks its logarithm's positivity condition. It then reports a familiar exact logarithm or an exact named logarithm with its defining exponential statement. An undefined argument, an excluded denominator input and a contextual exclusion are distinct reasons that must remain visible.

The output check uses the complete range and all permitted preimages, rather than asking whether your currently selected input happens to produce that output. The squared-input rule has two preimages for every real output; the canceled-inner rule has its whole original domain at output zero and none at another output. The bounded rule has a defined parent expression outside its contextual interval, but no contextual prediction there. Exact root or positive-power preimages can exist even when no control button lists them.

Use the logarithm definition and guarded laws to check these results, keeping each original domain. The following lesson will develop methods for exponential and logarithmic equations, including candidate checks against their original expressions. Here the foundation is the unique inverse meaning, complete range and preimage reasoning, and equality only on a justified shared domain. A simplification, estimate or plotted sample cannot supply an originally undefined value.

Worked examples

Try it yourself

1. Evaluate log₅(1/125), log_(1/5)(25), log₄(2) and log₉(27) exactly. State the real status of log₂(−1), log₂(0), log₁(1) and log_(−2)(8). Does a negative logarithmic value imply an invalid argument?

   Show the answer and reasoning

   The four values are −3, −2, 1/2 and 3/2. The last four expressions are not defined as standard real logarithms here. A negative logarithmic output can have a valid positive argument.

   Use 5⁻³ = 1/125, (1/5)⁻² = 25, 4^(1/2) = 2 and 9^(3/2) = 27. A positive-base exponential never gives a negative argument or zero. Base 1 cannot give a unique inverse exponent, and a negative base does not define a standard all-real exponential. The first two valid examples have negative exponents, not negative arguments.

2. Give the original real domains and simplified rules of P(t) = log₂(2ᵗ), Q(x) = 2^(log₂ x), H(x) = log₂(x²) and J(x) = 2 log₂ x. Compare H and J at x = −4 and explain whether their apparently related formulas define the same whole function.

   Show the answer and reasoning

   P(t) = t on all real t. Q(x) = x on x > 0. H(x) = 2 log₂|x| on x ≠ 0, while J(x) = 2 log₂ x on x > 0. H(−4) = 4 and J(−4) is undefined. H and J agree on x > 0 but have different original domains.

   The inner exponential in P is always positive, whereas the inner logarithm in Q first excludes nonpositive x. The square in H is positive at every nonzero x, including negative ones. Its magnitude |x| supplies the valid separate logarithm. J logs x itself and excludes every negative input. Neither an inverse cancellation nor a power-law rewriting enlarges an original domain.

3. For F(x) = log_(1/2)(5 − x) restricted to 1 ≤ x < 5, prove the complete range, find F(3), and give the sole permitted preimage of each output in the range. Explain why output −3 is not attained.

   Show the answer and reasoning

   The complete range is [−2, ∞), F(3) = −1 and each y ≥ −2 has sole permitted preimage x = 5 − 2^(−y). Output −3 would require x = −3, outside the stated domain.

   The argument runs through (0, 4], so it is always positive. The decreasing below-one logarithm at the largest argument 4 gives −2, and it grows without bound as the argument approaches zero positively. Inverting gives 5 − x = (1/2)ʸ = 2^(−y). For y ≥ −2 this positive argument is at most 4, placing x in [1, 5); it gives the proposed output and proves completeness. At x = 3 the argument is 2 = (1/2)⁻¹. For y = −3 the argument would be 8 and x = −3.

4. Compare P(x) = log₂((x + 2)(x − 2)) with Q(x) = log₂(x + 2) + log₂(x − 2). Find each original real domain, test x = −3 and give a logarithmic expansion that preserves P's entire original domain.

   Show the answer and reasoning

   P has domain x < −2 or x > 2; Q has domain x > 2. At x = −3, P = log₂ 5 and Q is undefined. On P's whole domain, P(x) = log₂|x + 2| + log₂|x − 2|.

   The product in P is x² − 4, positive outside [−2, 2]. Q instead requires both separate arguments to be positive, whose intersection is x > 2. At −3 the two factors are −1 and −5, whose positive product is 5. On both permitted branches the factors have the same sign, so |x + 2||x − 2| = (x + 2)(x − 2), and both magnitudes are positive. The guarded product law then gives the stated expansion without losing the negative-input branch.

5. For T(x, z) = log₂(√x/z²), determine the original real domain, expand using separate logarithms without losing negative permitted z, and evaluate T(64, −2). Does the existence of √0 permit x = 0 here?

   Show the answer and reasoning

   The domain is x > 0 and z ≠ 0. On that domain, T(x, z) = (1/2) log₂ x − 2 log₂|z|. T(64, −2) = 1. Input x = 0 remains excluded.

   The square root requires x ≥ 0 and the quotient requires z ≠ 0; the logarithm requires its quotient to be positive, so x must actually be positive. On this domain both √x and z² are positive, allowing the quotient law. The root power law gives half the log of x, and z² = |z|² gives twice the log of its positive magnitude. At (64, −2), √64/z² = 8/4 = 2. At zero x the quotient would be zero, whose real logarithm is undefined.

6. Use base 2 to evaluate log_(1/4)(8) and log₄(1/8) exactly. Express log₃(5) using natural logarithms, state every condition needed for the change-of-base formula and explain why the denominator is nonzero.

   Show the answer and reasoning

   Both exact values are −3/2. The exact expression log₃(5) is ln 5/ln 3. In general require x > 0, b > 0, b ≠ 1, c > 0 and c ≠ 1 in log_b x = log_c x/log_c b; its denominator is nonzero because b ≠ 1.

   The first ratio is log₂ 8/log₂(1/4) = 3/(−2); the second is log₂(1/8)/log₂ 4 = (−3)/2. The base used for the numerator and denominator must be the same valid c. The original positive base b is itself a valid argument for log_c; its logarithm equals zero only when b = 1, which the formula excludes. Natural logarithm uses the valid standard base e, and the exact ratio does not require a rounded decimal.

7. A fictional scale is S(A) = log₂(A/50) for numerical A in litres, with a fixed 50-litre reference and permitted amounts 25 ≤ A ≤ 200 litres. Prove its complete range, find the preimage of scale output −1/2 and check output 3. Starting from 100 litres, a stipulated 20% increase followed by a 20% decrease gives 120 then 96 litres. What is the exact total change in S, and is it zero?

   Show the answer and reasoning

   The complete range is [−1, 2]. Output −1/2 has sole permitted preimage 50/√2 = 25√2 litres. Output 3 would need 400 litres and is not permitted. The total scale change from 100 to 96 litres is log₂(24/25), a strictly negative value rather than zero.

   The argument interval is [1/2, 4], and every y in [−1, 2] has the permitted preimage A = 50·2ʸ litres; strict increase excludes other outputs. Since 1 < √2 < 2, 25√2 lies between 25 and 50 litres. The two percentage factors multiply to (6/5)(4/5) = 24/25, with both intermediate amounts permitted. The quotient law gives S(96) − S(100) = log₂(96/100) = log₂(24/25). Its positive argument is below 1, so its base-2 logarithm is negative. Opposite percentages do not undo one another.

What to carry forward

• For b > 0, b ≠ 1 and x > 0, log_b x is the unique real exponent y satisfying bʸ = x. An exact named logarithm can be a complete answer even without a familiar fractional value.
• The full real logarithm has positive arguments and all real outputs. Zero or negative arguments are invalid here; negative logarithmic values can be valid, and bases between zero and one are permitted.
• A parent logarithm is increasing for b > 1 and decreasing for 0 < b < 1. Prove a complete range by checking output restrictions and supplying every permitted preimage, including contextual endpoints.
• The identities log_b(bᵗ) = t and b^(log_b x) = x have different starting domains. Inner roots, denominators and contextual restrictions remain in force after cancellation or simplification.
• Product and quotient laws require positive separate arguments on a shared original domain. A positive product or quotient of negative parts does not supply their separate real logarithms, and there is no logarithm addition law inside an argument.
• For U > 0 and real r, log_b(Uʳ) = r log_b U. On x ≠ 0, log_b(x²) = 2 log_b|x|; changing it to 2 log_b x would lose the negative-input branch. Change of base requires two valid bases and a nonzero log of the original base.
• A stipulated logarithmic ratio scale uses positive dimensionless arguments and a fixed positive reference. Equal multiplicative factors become equal added scale steps only where both compared original inputs are permitted.