Logarithms

01 / UNDERSTAND & APPLY

1. A logarithm asks for an exponent

For a real base a with a>0 and a≠1, logₐx is the unique real exponent y such that aʸ=x. The argument x must be positive. The restrictions are structural: positive bases produce positive outputs; base 1 produces only the output 1 and therefore cannot have an inverse assigning a unique exponent.

An argument need not exceed 1. For example, log₃(1/9)=−2 because 3⁻²=1/9. A logarithm may therefore be negative, zero or positive even though its argument must always be positive. Also logₐ1=0 and logₐa=1 for every valid base.

The notation ln x means the logarithm with base e. Common logarithm means base 10, often written log x when that convention is stated. Never infer a base from a context that has not specified it, especially when different texts use log for different bases. Identities involving ratios of logarithms are base-independent only under their precise change-of-base conditions.

The exponential and logarithmic functions undo each other on their domains: a^(logₐx)=x for x>0, and logₐ(aᵗ)=t for every real t. The first identity starts with a positive input; the second can start with any real exponent. Their different domains matter in nested expressions.

02 / UNDERSTAND & APPLY

2. Build the full domain before any simplification

Every real logarithm has two kinds of restrictions: its argument is positive, and its base is positive but not 1. With a constant valid base, only the argument restriction varies. With a variable base, all three conditions must be carried through the solution.

Nested logarithms impose restrictions from the inside out and then from the outside. The inner logarithm must exist, but its value must also be a positive argument for the outer logarithm. For ln(ln x), the inner domain x>0 is not enough; ln x>0 strengthens it to x>1.

A logarithm in a denominator adds a nonzero condition on its value. Since logₐu=0 exactly when u=1 for a valid base, a denominator logₐu excludes u=1 as well as nonpositive u. If that logarithm is itself the base of another logarithm, it must be positive and unequal to 1 instead.

Combining expressions can produce a formula defined at inputs where the original expression was undefined. Such inputs do not become valid merely because the new formula accepts them. Keep the original domain alongside every equivalent algebraic step, and test final candidates in the original question.

03 / UNDERSTAND & APPLY

3. Product, quotient and power laws come from exponents

Let u,v>0 and write u=aʳ, v=aˢ using a valid base a. Their product is a^(r+s), so logₐ(uv)=r+s=logₐu+logₐv. Their quotient is a^(r−s), giving the quotient law. These derivations explain both the algebra and the positivity assumptions.

For u>0 and real k, define uᵏ through the usual positive-base real power. Then uᵏ=a^(kr), giving logₐ(uᵏ)=k logₐu. If u is allowed to be negative and k is an even integer, the left side may still exist while logₐu does not. In particular, ln(x²)=2ln|x| for x≠0; writing 2ln x restricts it to x>0.

There is no general rule turning logₐ(u+v) into logₐu+logₐv. The latter is the logarithm of a product, not a sum. For example, log₂(2+2)=2 while log₂2+log₂2=2 happens to agree, but replacing both terms by 3 breaks the proposed identity. One successful numerical case is not a proof.

Expanding and combining are useful in different directions. Expansion exposes factors and exponents. Combining can reduce an equation to one logarithm, but requires each original argument to be positive. The condensed expression logₐ(uv) by itself also accepts two negative factors, whereas the sum of their real logarithms does not.

04 / UNDERSTAND & APPLY

4. Changing base is a ratio of exponents

Let y=logₐx, so aʸ=x. Take logarithms in another valid base b: y log_b a=log_b x. Since a≠1, the factor log_b a is nonzero and division gives logₐx=log_b x/log_b a. The same formula with natural logarithms is often the safest way to analyze a variable base.

For a,b>0 with a≠1 and b≠1, logₐb and log_b a are reciprocals. More generally logₐb·log_b c=logₐc when every displayed logarithm exists. The cancellation is a consequence of the ratio formula, not permission to cancel arbitrary printed “log” symbols.

The identity log_(aʳ)(xˢ)=(s/r)logₐx holds for a>0, a≠1, x>0 and r≠0. The new base aʳ must remain valid; r=0 would make it 1. The exponent s may be zero because the resulting argument 1 is allowed.

When a variable appears in a base, record the excluded value 1 before introducing a logarithmic substitution. For instance, t=log₂x with x>0 and x≠1 gives t∈ℝ∖{0}. Then log_x2=1/t, and a rational equation in t can be solved without losing that restriction.

05 / UNDERSTAND & APPLY

5. Solve equations on the original admissible set

For a fixed valid base, logₐu=logₐv is equivalent to u=v provided both arguments are positive. Likewise logₐu=c is equivalent to u=aᶜ. If the logarithms have different bases, convert them to a common base before comparing arguments.

A sum of logarithms often produces a polynomial equation after combining. The polynomial may have roots outside the logarithm domain, so solving it is only the candidate-generation stage. Domain rejection is not an arbitrary final adjustment; it restores the conditions that made the logarithmic transformation valid.

For expressions polynomial in logₐx, use t=logₐx. Because a>0 and a≠1, this substitution maps x>0 bijectively onto all real t. Each real t solution produces exactly one positive x=aᵗ. Additional logarithms or denominators may impose further restrictions on t.

Exponentiating both sides with a fixed valid base preserves equality because the exponential is injective. Squaring both sides, by contrast, may introduce a sign ambiguity. Keep track of which transformations are reversible rather than assuming that every algebraic manipulation preserves the solution set automatically.

06 / UNDERSTAND & APPLY

6. The base controls the inequality direction

For a>1, the function logₐx is strictly increasing on x>0, so comparing logarithms preserves the order of their arguments. For 0<a<1, it is strictly decreasing, so the order reverses. These are whole-domain monotonicity facts, not merely sign rules at a few points.

Thus logₐu<c becomes u<aᶜ for a>1, but u>aᶜ for 0<a<1. The argument restriction u>0 remains in either case. Replacing a logarithmic inequality by an algebraic one without carrying its domain can include intervals where the original statement was not defined.

The sign of logₐx follows by comparing x with 1, because logₐ1=0. For a>1 it has the sign of x−1, while for 0<a<1 the sign is reversed. If a logarithm appears as a factor or denominator, these sign intervals must be included in the sign chart.

For inequalities involving two logarithms with the same fixed base, impose positivity on both arguments first. Then compare the arguments with the appropriate direction. A factor that is positive on the domain can be multiplied safely; otherwise split into its positive and negative cases.

07 / UNDERSTAND & APPLY

7. Split at every base transition

For log_xu(x), the base requires x>0 and x≠1. The intervals 0<x<1 and x>1 have opposite monotonicity, so a single comparison rule cannot cover both. Also impose u(x)>0 before solving either branch.

Alternatively write log_xu=ln u/ln x. The denominator ln x is negative on (0,1) and positive on (1,∞). Multiplying an inequality by it is legitimate only after that sign split. The ratio form and the base-monotonicity form are two descriptions of the same reasoning.

If the base is a more complicated expression b(x), split where b(x)=0 or b(x)=1 and keep only b(x)>0. Argument zeros, poles and denominator zeros may add further boundary points. A final interval union should state open or closed endpoints according to the original expression, not only the transformed inequality.

08 / UNDERSTAND & APPLY

8. Nested domains and parameter feasibility

Nested logarithmic equations are often easiest to unwind from the outside after establishing that each inner stage is legal. For log₂(log₃x)=2, the outer equation requires log₃x=4, hence x=81. The solution satisfies the stronger outer-domain condition log₃x>0, which here is x>1.

For a parameter equation, determine whether the transformed real values can occur. The expression t+1/t for t>0 has minimum 2, since (t−1)²/t≥0. For t<0 it is at most −2 by substituting t=−s. Therefore its range over nonzero real t is (−∞,−2]∪[2,∞).

Applied to log₂x+log_x2=k, this shows there are no positive-domain solutions when −2<k<2, one solution at each boundary k=±2, and two solutions for |k|>2. At k=2, t=1 gives x=2; at k=−2, t=−1 gives x=1/2. This analysis also checks the discriminant k²−4 of the quadratic t²−kt+1=0.

Logarithmic scales describe multiplicative ratios through additive differences. If a quantity is explicitly defined by L=log₁₀(Q/Q₀) for positive Q,Q₀, a change of 2 in L represents a factor 100 in Q. Always use the stated physical definition; different scientific scales can include additional constants or measure different quantities.

09 / UNDERSTAND & APPLY

9. Original practice with domain checks

Write the admissible domain before solving. When a base lies below 1 or varies with x, explain the direction of each inequality transformation.

10 / UNDERSTAND & APPLY

10. Keep legality, equality and order separate

The reliable order of work is domain first, algebra second, verification last. Identify every argument, every base and every denominator. Then choose a logarithm law, a change of base or a substitution that simplifies the structure without discarding the original admissible set.

The recurring mistakes are allowing zero or negative arguments, allowing base 1, splitting a logarithm of a sum, dropping absolute values after an even power, comparing arguments with different bases, and multiplying an inequality by ln x without knowing its sign. Each mistake has a precise domain or monotonicity diagnosis.

For final checks, convert a simple logarithm back to an exponential, substitute each proposed equation root, and test an interior point from each inequality interval. Preserve exact values such as √2 or 3^(1/2) until an approximation is requested. Numerical rounding near a strict endpoint can obscure whether that endpoint belongs.

A logarithm turns multiplication into addition because it measures exponents. Remembering that origin makes the rules coherent: products add exponents, quotients subtract them, powers multiply them, and a changed base rescales the unit in which an exponent is measured.