Exponential and logarithmic equations

A solution must make the original equation defined and true

A real solution is an input at which both original sides are defined and have equal values. Start by recording every logarithm's positive-argument condition, every inner denominator or root condition and any stated contextual restriction. Their intersection is the original admissible domain D. Solving a simpler equation can produce candidate inputs; a candidate outside D is not a solution, even if the simplified equation is true there.

A step is reversible on D when the original and transformed equations are true at exactly the same inputs in D. Adding the same defined quantity to both sides or multiplying both sides by a known nonzero quantity preserves equality there. Dividing by a variable expression needs its nonzero condition; squaring can lose sign information. A chain of justified equivalences gives a completeness proof, while a necessary-condition step requires a later check and an argument that every original solution was considered.

Separate three outcomes. The equation 2ˣ = −1 is defined at every real x but has no real solution because its left side is positive. The equation log₂ x = log₂(−x) has no admissible real input, because x > 0 and −x > 0 cannot both hold. The equation log₂((x − 2)/(x − 2)) = 0 is an identity on x ≠ 2: every original admissible input is a solution, while x = 2 stays undefined. An empty solution set, an empty original domain and a whole-domain identity have different reasons.

Equal powers of one valid base have equal exponents

For b > 0 and b ≠ 1, the standard real exponential is one-to-one: b^U = b^V is equivalent to U = V whenever the original exponent expressions U and V are defined. This is true for b > 1 and for 0 < b < 1. Equality does not require reversing a sign for a decreasing base; that issue belongs to inequalities. Preserve any restrictions already needed to evaluate U and V.

For example, 4^(x + 1) = 8^(x − 1) becomes 2^(2x + 2) = 2^(3x − 3), using positive-base power laws. Equating exponents gives x = 5. Substitution gives 4⁶ = 8⁴ = 4096, and the reversible steps show it is the only real solution. Roots and reciprocals can also reveal a common base: (1/2)^(x − 1) = 4 becomes 2^(1 − x) = 2² and gives x = −1.

The base condition matters. The equation 1ˣ = 1^(2x) holds for every real x, whereas setting x = 2x would incorrectly reduce its solutions to zero. A negative fixed base does not provide the standard all-real exponential used by this method. Rewriting a power to a positive common base must be a valid identity on the original domain, rather than a convenient resemblance between selected integer powers.

Isolate the exponential before taking its logarithm

Consider a·b^(mx + c) + d = Y with real constants, a valid base b and a ≠ 0. Subtract d and divide by a to get b^(mx + c) = q, where q = (Y − d)/a. If q ≤ 0 there is no real solution: a positive-base exponential is always positive, so taking log_b q is not a permitted step. If q > 0, the inverse definition gives mx + c = log_b q.

When m ≠ 0, the sole unrestricted real solution is x = (log_b q − c)/m. Check any original context bounds separately. For 3·2^(2x − 1) + 1 = 10, isolation gives 2^(2x − 1) = 3, hence x = (1 + log₂ 3)/2. This exact value lies between 1 and 3/2 and gives the original left side 3·3 + 1 = 10. The isolated exponential with its nonzero affine exponent coefficient is one-to-one, proving uniqueness.

Treat constant cases before dividing. If a = 0, the given affine-exponent expression is defined for every real x and its left side equals d; equality holds everywhere or nowhere according as Y = d or Y ≠ d. If a ≠ 0 but m = 0, the left side is the constant a·bᶜ + d and the same constant comparison applies. A contextual domain restricts every 'everywhere' answer to its permitted inputs. These conclusions concern the stated, defined exponential expression; multiplying an undefined logarithm by zero would not repair its original domain.

Different bases can give a linear equation in the exponent

When both complete sides are positive, applying the same valid logarithm to them is reversible because that logarithm is one-to-one. For 3^(2x + 1) = 5ˣ, both sides are positive at every real x. Natural logarithms give (2x + 1) ln 3 = x ln 5, so x(2 ln 3 − ln 5) = −ln 3. Since 2 ln 3 − ln 5 = ln(9/5) > 0, division gives the sole real solution x = −ln 3/ln(9/5). The nonzero denominator is justified, not assumed.

A logarithm of a power can bring its real exponent down only under the positive-base power-law guards. A logarithm does not distribute over an added term: ln(U + V) is not ln U + ln V. Isolate a single exponential when possible before using logs; the sum 2ˣ + 3ˣ does not become a linear expression in x merely by logging it. We use methods that genuinely reduce each particular equation, rather than claiming a universal elementary formula.

Do not apply a real logarithm to a negative or zero whole side. An equation's original sides can nevertheless be negative and equal, or a logarithmic equation can have a negative logarithmic value. For example, log₂ x = −2 is valid and gives x = 1/4 by the inverse definition; taking a further real log of its right side would be invalid. Named logarithms and exact ratios of them remain complete exact solutions without rounding.

A polynomial in an exponential still needs positive substitution values

For a valid fixed base b, set u = bˣ. This is a one-to-one correspondence between all real x and all positive u, with inverse x = log_b u. The extra condition u > 0 is essential: a real root of a polynomial in u need not be an output of the exponential. Express every occurrence of the exponential through this same u, solve the resulting polynomial completely and then keep all and only its positive roots.

For 2^(2x) − 5·2ˣ + 4 = 0, the positive-base identity 2^(2x) = (2ˣ)² gives u² − 5u + 4 = (u − 1)(u − 4) = 0. Both roots 1 and 4 are positive, so they give x = 0 and x = 2. These are two genuine solutions; the exponential's own one-to-one property does not imply that a polynomial in its output is one-to-one. Factoring the complete quadratic and the substitution's bijection prove that there are no other real solutions.

For 2^(2x) + 2ˣ − 2 = 0, the polynomial is u² + u − 2 = (u − 1)(u + 2). Its root u = −2 is inadmissible, while u = 1 gives the sole real solution x = 0. There is no real candidate log₂(−2). Similarly, u = 0 would be rejected: no finite real x makes a positive-base exponential zero. Repeated positive polynomial roots give the same input once, not extra distinct solutions.

A reciprocal substitution is safe because the exponential is nonzero

If an equation contains bˣ and b⁻ˣ, the same substitution u = bˣ > 0 turns them into u and 1/u. For 3ˣ + 3⁻ˣ = 10/3, the equation is u + 1/u = 10/3. Multiplying by 3u is reversible on u > 0, giving 3u² − 10u + 3 = (3u − 1)(u − 3) = 0. Both positive roots survive: u = 1/3 gives x = −1 and u = 3 gives x = 1.

The fact that multiplying by u is valid here comes from the substitution's positivity, not from a blanket permission to clear any variable denominator. In a different original rational expression, record each denominator exclusion first. If a reduced polynomial has a root at an excluded denominator value, it must be filtered out. For the exponential substitution, zero is already excluded throughout the entire chain.

A completed finite-root argument lists every root of the reduced polynomial, filters by u > 0 and the original domain, maps each retained u back through the unique logarithm and checks the original equation. A failed candidate check alone does not prove that an unrelated candidate list is complete. In the example, the factorisation enumerates every polynomial root and the reversible correspondence covers every real x, which is why the two checked solutions form the full set.

An isolated logarithm can be rewritten as a positive target

For a valid base b, log_b F(x) = r is equivalent to F(x) = bʳ on inputs where the original inner expression F is defined. The target bʳ is positive, so any retained solution of that target equation automatically makes the outer logarithm's argument positive. Still keep inner denominator and root restrictions: a reduced formula for F can have values at inputs where its original expression does not exist.

For log₂√(x − 1) = −1, first require the root to exist and its value to be positive, giving x > 1. Inverting gives √(x − 1) = 1/2. Both sides are nonnegative, so squaring is reversible here and gives x − 1 = 1/4, hence x = 5/4. The original check gives log₂(1/2) = −1. Positivity makes this particular squaring safe; squaring a general equation without its sign conditions can admit unwanted candidates.

An inverse rewrite may also produce a contradiction or an identity. For log₂((x − 2)/(x − 2)) = 0, the inner denominator first excludes x = 2. Its quotient is 1 everywhere else, giving every x ≠ 2 as a solution. For the same left side equal to 1, it would require the quotient to be 2, so there are no solutions on that nonempty original domain. Simplifying to a constant never inserts the missing denominator input.

Equal logarithms need two defined positive arguments

For b > 0 and b ≠ 1, log_b U(x) = log_b V(x) is equivalent to U(x) = V(x) on the original domain where both inner expressions are defined and U(x) > 0 and V(x) > 0. This follows from the logarithm's one-to-one property and applies equally to bases between zero and one. Equal values do not need an order reversal. Preserve the intersection of both positive-argument conditions before solving the simpler equation.

For log₂(x − 1) = log₂(2x + 1), the original conditions give x > 1. Equating arguments gives x − 1 = 2x + 1, hence candidate x = −2. Its two arguments would both be −3, which makes the original real logarithms undefined. The original equation has no solution, even though its reduced linear equation has a real root. Every possible original solution would have to be that root, so rejecting it proves completeness of the empty answer.

The bases must be the same for the direct argument-equality step. Log₂ 4 = 2 and log₄ 16 = 2 are equal logarithmic values with different arguments; different bases do not imply that their arguments are equal. Use a valid change-of-base formula or the inverse definitions when the bases differ, and justify any division by a logarithmic coefficient. Do not discard original argument conditions when changing notation.

Combining logs cannot enlarge the original domain

A sum of logarithms with the same valid base can become the logarithm of a product only where every separate argument is positive. A difference can become the logarithm of a quotient under the same requirement. For log₂(x − 1) + log₂(x + 1) = 3, the original intersection is x > 1. On that domain, condensation gives log₂(x² − 1) = 3, hence x² − 1 = 8 and algebraic candidates x = ±3.

The candidate x = 3 gives the original arguments 2 and 4 and the sum 1 + 2 = 3. Candidate x = −3 gives negative arguments −4 and −2; it is not a real solution. Its positive product would satisfy the condensed logarithm, illustrating that the condensed expression has a larger natural domain if the original restriction is dropped. Solving the quadratic enumerates all possible original solutions; preserving x > 1 leaves precisely x = 3.

A power-law condensation needs the same discipline. The equation 2 log₂ x = 4 has domain x > 0 and sole solution x = 4. Condensing to log₂(x²) = 4 while retaining that domain gives the same answer. If you instead treat the condensed formula on its whole natural domain x ≠ 0, it also admits x = −4 and describes a different equation. Replacing a separate argument by its absolute value changes the starting expression unless that equality has already been justified on the original domain.

An exact time must still belong to the supplied model

A stipulated continuous model A(t) = A₀b^(t/τ), with A₀ > 0, b > 0, b ≠ 1 and τ > 0, reaches a positive target Y in its full real-time extension at the unique numerical time t = τ log_b(Y/A₀). The ratio is dimensionless, and t and τ use the same time unit. If the context allows only 0 ≤ t ≤ T, check that interval or the complete contextual range before reporting an attained target. A mathematical extension need not be a permitted past or future observation.

For an offset rule A(t) = L + Cb^(t/τ) with C ≠ 0, isolate b^(t/τ) = (Y − L)/C. This dimensionless ratio must be positive. The offset level Y = L would need an exponential output zero and is not attained at finite time. If C = 0 the model is constant L, so the target is either attained at every permitted time or never. The factor acts on A − L, not generally on the total A; omitting the offset changes the equation.

A discrete stage model has integer n, rather than every real t. Solving A₀bⁿ = Y as a real equation supplies a candidate n = log_b(Y/A₀), but it is a stage solution only if that candidate is an allowed integer stage. Joining plotted stage points or rounding a noninteger candidate cannot create an exact stage when the target occurs. These calculations use supplied ideal rules and domains; algebra alone does not establish that an observed process will continue under unchanged conditions.

Check one candidate without confusing it with the whole answer

The explorer presents seven parameterised cases. Select the real parameter r and an input candidate independently: r is an exponent, a target or a polynomial coefficient according to the selected equation, rather than one interchangeable quantity. A candidate check first asks whether the input belongs to the original domain and then whether the original sides are equal. Read a defined false equation separately from an undefined original expression, and read a contextual exclusion separately from a defined formal extension. Exact powers, positive roots and named logarithms replace rounding-based equality verdicts.

For the substitution family u² − 3u + r = 0 with u = 2ˣ > 0, the discriminant is 9 − 4r. If r < 0 the roots have opposite signs and only the positive one survives; at r = 0 the roots are 3 and the excluded zero. For 0 < r < 9/4 both roots are positive because their positive sum is 3 and positive product is r. At r = 9/4 the repeated root 3/2 gives one distinct solution, and at r > 9/4 there are no real polynomial roots. Each retained positive u has the unique exact preimage log₂ u.

For the log-sum case the original domain is x > 1 and the sole solution is √(1 + 2ʳ), greater than 1 for every real r; its negative algebraic partner makes the original separate arguments negative. For the log-equality case the original domain is 1 < x < r, empty when r ≤ 1. When r > 1 its midpoint candidate (r + 1)/2 is inside that domain and solves the equation. The canceled-inner case instead has nonempty domain x ≠ 2 for every r: at r = 0 it is an identity, and otherwise it is defined but false at every permitted input.

The bounded amount model is B(t) = 8·2^(−t) litres with numerical t in hours, a one-hour halving period and 0 ≤ t ≤ 2. Its complete contextual range is [2, 8] litres. At target r = 1 litre the formal time is 3 hours, outside that context, whereas a nonpositive target has no finite real time even in the extension. For every target in [2, 8] litres the unique time is log₂(8/r) hours. A short candidate control list is neither the original domain nor a proof of completeness; use the complete branches and range conditions in the table.

Worked examples

Try it yourself

1. Solve 4^(x + 1) = 8^(x − 1) and (1/4)^(2x + 1) = 8 exactly. Explain why the decreasing base in the second equation does not require reversing an equality, and check that the common-base method covers every real solution.

   Show the answer and reasoning

   The first equation has sole solution x = 5. The second has sole solution x = −5/4. Equal powers of any valid base imply equal exponents, whether the base is above or below one.

   Rewrite the first equation as 2^(2x + 2) = 2^(3x − 3), giving 2x + 2 = 3x − 3 and x = 5. Rewrite the second as 2^(−4x − 2) = 2³, giving −4x − 2 = 3 and x = −5/4. At the first root both sides are 4096; at the second, exponent 2x + 1 is −3/2 and (1/4)^(−3/2) = 8. All original expressions are defined on the reals and the valid common-base steps are reversible, so there are no further roots.

2. Solve 5 − 2·3^(x − 1) = 1, and then solve 5 − 2·3^(x − 1) = 7. Also classify the equations 0·3ˣ + 5 = 5 and 0·3ˣ + 5 = 7 on all real inputs. State why no division by a zero coefficient is needed.

   Show the answer and reasoning

   The first equation has sole solution x = 1 + log₃ 2. The second has no real solution. The third is an identity on all real x, and the fourth has no solution.

   The first isolates 3^(x − 1) = 2, a positive target with unique inverse exponent. Substitution gives 5 − 2·2 = 1. The second would require 3^(x − 1) = −1, impossible for the positive-base exponential. For the zero-coefficient equations, 3ˣ is defined at every real input and the whole left side is the constant 5. Compare constants directly instead of dividing by zero; the two constant equations respectively agree and disagree everywhere.

3. Solve 2^(3x) = 7^(x − 1) exactly using natural logarithms. Justify the denominator's nonzero value and explain why logging both sides preserves all real solutions.

   Show the answer and reasoning

   The sole real solution is x = −ln 7/ln(8/7), a negative exact number.

   Both original sides are positive on all real x, so the same valid natural logarithm is reversible. Its power law gives 3x ln 2 = (x − 1) ln 7. Hence x(ln 8 − ln 7) = −ln 7 and x = −ln 7/ln(8/7). The denominator is positive because 8/7 > 1 and ln is increasing with ln 1 = 0; it is therefore nonzero. The resulting linear equation has a unique root and the preceding steps are equivalences, proving completeness without a rounded decimal.

4. Solve 2^(2x) − 6·2ˣ + 8 = 0 and 3ˣ + 3⁻ˣ = 10/3 over the reals. Record the positive-substitution conditions, retain every permitted branch and explain why clearing the reciprocal denominator in the second equation is valid.

   Show the answer and reasoning

   The first complete solution set is {1, 2}. The second is {−1, 1}.

   For the first, u = 2ˣ > 0 gives u² − 6u + 8 = (u − 2)(u − 4) = 0. Both positive roots give x = log₂ 2 = 1 and x = log₂ 4 = 2. For the second, u = 3ˣ > 0 gives u + 1/u = 10/3; multiplying by nonzero 3u gives (3u − 1)(u − 3) = 0. Its positive roots 1/3 and 3 give x = −1 and x = 1. Each factorisation is complete and each positive root has exactly one exponential preimage. Direct original checks give 4 − 12 + 8 = 0 and 16 − 24 + 8 = 0 in the first, and 3 + 1/3 = 10/3 in either order in the second.

5. Solve log₅(x² − 4) = log₅(3x). Record the full original real domain before equating arguments, list every algebraic candidate and check the permitted answer. Is equality of the reduced arguments at a rejected input enough?

   Show the answer and reasoning

   The original domain is x > 2. The algebraic candidates are 4 and −1; the complete original solution set is {4}. Reduced-argument equality does not make an undefined original logarithm valid.

   The first argument requires x < −2 or x > 2, while 3x > 0 requires x > 0; their intersection is x > 2. On this domain, same-base injectivity gives x² − 4 = 3x, or (x − 4)(x + 1) = 0. At x = 4 both original arguments are 12 and their logarithms are equal. Candidate −1 is outside the domain and gives both arguments −3, so neither real logarithm exists. Every original solution must appear among the two complete quadratic roots, and exactly one is permitted.

6. Solve log₂(x − 3) − log₂(x + 1) = −1 and log₂(x − 3) + log₂(x + 3) = 4. State each original domain, justify the quotient or product law there and check every reduced algebraic candidate.

   Show the answer and reasoning

   Both original domains are x > 3. The first complete solution set is {7}; the second is {5}, with algebraic candidate −5 rejected.

   For the first, positive separate arguments allow log₂((x − 3)/(x + 1)) = −1, so the ratio is 1/2 and 2x − 6 = x + 1 gives x = 7. The original check is log₂ 4 − log₂ 8 = 2 − 3 = −1. For the second, condensation gives log₂(x² − 9) = 4, hence x² = 25 and candidates ±5. Only 5 exceeds 3, and its original check is log₂ 2 + log₂ 8 = 1 + 3 = 4. At −5 the separate arguments are −8 and −2 and their logarithms are undefined, although their product is positive. Each reversible chain on the original domain exhausts its candidates.

7. An ideal continuous amount model is B(t) = 80·2^(−t/5) litres with numerical t in hours and 0 ≤ t ≤ 10. Find its complete range and solve B(t) = 40, B(t) = 15 and B(t) = 0 within that domain. A separate stage-only rule C(n) = 80·2^(−n) litres allows n = 0, 1, 2. Is 30 litres attained by C, and when is it attained by B?

   Show the answer and reasoning

   B has complete range [20, 80] litres. Its sole time for 40 litres is 5 hours; 15 litres and zero have no permitted time. The stage-only range is {80, 40, 20} litres, so C never gives 30. B gives 30 litres at the sole time 5 log₂(8/3) hours, between 5 and 10 hours.

   The continuous exponential is strictly decreasing from 80 to 20 and is continuous, giving every intervening amount. For 40 litres the ratio is 1/2 and t = 5. Target 15 would need t = 5 log₂(16/3), greater than 10 because 16/3 > 4; it lies outside the time interval. Zero would require a zero exponential output and is never attained at finite time. The three integer stages explicitly give 80, 40 and 20. For the continuous target 30, the dimensionless ratio 30/80 = 3/8 gives t = 5 log₂(8/3), with 2 < 8/3 < 4 proving 5 < t < 10. A noninteger stage preimage cannot create a permitted integer stage.

What to carry forward

• Record the intersection of every original logarithm, denominator, root and context condition. A solution makes the original equation both defined and true; a reduced algebraic candidate alone is not enough.
• For b > 0 and b ≠ 1, equal powers of the same base have equal exponents. The method works for bases between zero and one, while base one requires a separate constant comparison.
• Isolate an exponential and require its target to be positive before taking its logarithm. Check zero coefficients and constant exponent cases before dividing, and retain exact named logarithms when a familiar power is unavailable.
• A substitution u = bˣ requires u > 0 and has unique inverse x = log_b u. Factor the reduced polynomial completely, retain every positive branch and reject zero or negative substitution roots.
• Same-base logarithm equality permits equating arguments only where both original arguments are defined and positive. Product, quotient and power-law condensation keep the same original restrictions and can otherwise admit extra inputs.
• Prove completeness through reversible steps and a complete reduced root list, then check retained candidates in the original equation. Distinguish an identity on its original domain, a defined equation with no solution and an equation with no admissible inputs.
• In a supplied model, subtract an offset before forming a positive dimensionless ratio, keep time and period units consistent and filter exact inverse times through the contextual interval or integer-stage domain.