Two methods and one decision. If both sides can be written as powers of the same base, match the exponents and the answer is exact. If they cannot, take a logarithm of both sides: the power law is the only tool in this course that moves a variable out of an exponent.
Core ideas
Everything here rests on one property of the exponential function and one law of logarithms.
For b > 0 with b ≠ 1, b^u = b^v forces u = v. That is what licenses cancelling the base and comparing exponents. It is the same property that gave the exponential an inverse in “Logarithmic Functions and Their Graphs”, and it fails at b = 1, where 1^u = 1^v for every pair of exponents.
log(M^p) = p · log M takes the exponent and writes it in front. Nothing else in algebra does this: dividing, factoring and squaring all leave x exactly where it was. When the same-base method fails, this law is the only route left.
Taking log of both sides of 5 · 2^x = 40 is legal and gives log 5 + x · log 2 = log 40, which does solve. It is also the line where log(5 · 2^x) gets expanded wrongly. Dividing by 5 first leaves 2^x = 8, where the power law has a clean target and there is nothing to invent.
The material
The same-base method, the logarithm method, the substitution that hides a quadratic, and the equations with no solution at all.
Every equation before this course put the unknown in the base: x² = 9, x³ − x = 0. 3^(2x − 1) = 81 puts it in the exponent, and no amount of dividing, factoring or root-taking will move it. Two methods do.
For b > 0 with b ≠ 1: b^u = b^v if and only if u = v.
The forward direction is what solves equations. The reverse direction is why substituting an answer back in works. Both statements fail for b = 1, which is why that base is excluded everywhere in this course.
3^(2x − 1) = 81. Since 81 = 3^4, both sides are powers of 3: 3^(2x − 1) = 3^4. One-to-one gives 2x − 1 = 4, so x = 5/2. The answer is exact and no logarithm was needed. Check: 2 · (5/2) − 1 = 4 and 3^4 = 81.
That method needs both sides to be recognisable powers of one base. 5^x = 17 is not such an equation. 17 lies between 5^1 = 5 and 5^2 = 25, and it is not a rational power of 5 at all: 5^(p/q) = 17 would mean 5^p = 17^q, which no positive integers satisfy, since one side is divisible by 5 and the other is not.
If M = N and both are positive, then log_c(M) = log_c(N) for every legal base c; and conversely, log_c(M) = log_c(N) forces M = N.
So a logarithm may be applied to both sides of an equation exactly as 5 may be added to both sides. The two-way statement matters: it is what guarantees the new equation has the same solutions as the old one, so nothing is gained or lost by the step.
The exact answer is the quotient ln 17 / ln 5, which is log_5(17) written in a base that can be computed — “Change of Base”, doing the job it was derived for. The decimal is a rounding of it, and any later step should use the quotient.
When the two sides have different bases and both carry the unknown, take a logarithm anyway and then collect. The equation that results is linear in x, with logarithms as its coefficients.
e^(2x) − e^x − 6 = 0 contains only one exponential, twice: e^(2x) = (e^x)². Substituting u = e^x turns it into u² − u − 6 = 0, which factors as (u − 3)(u + 2) = 0. So u = 3 or u = −2.
Now undo the substitution, and one candidate dies. u = e^x is positive for every real x, so u = −2 is impossible — not because of a domain restriction on a logarithm, but because the range of the exponential is (0, ∞). From u = 3, x = ln 3 ≈ 1.098612, and that is the whole solution set.
The same range fact settles two shapes on sight. b^x = 0 has no solution: the horizontal asymptote y = 0 from “Exponential Functions” is approached and never reached. b^x = k with k < 0 has no solution either. In both cases the algebra will not object — ln(−8) simply does not exist — so the check has to be made before the logarithm, not after.
Interactive
Match the powers when you can; take logs when you cannot
Method
Four steps. The decision in step two is the one that determines whether the answer is exact.
Get the power alone on one side. Constants added to it move first, then coefficients multiplying it — the same order as in a linear equation, and for the same reason.
If it is, rewrite both sides over that base and set the exponents equal. The answer is exact, and no rounding enters the problem.
Any base works; use ln or log because they can be evaluated. The power law brings the whole exponent down in front as a factor, so bracket it before you write it: (2x − 1) · ln 3, never 2x − 1 · ln 3.
After the logarithm the equation is linear in x with numeric coefficients. If the isolated exponential equalled 0 or a negative number, stop earlier: there is no real solution to find.
The same four steps each time. What differs is where they stop.
Equation (2) is the one worth studying. Two isolating steps happen before anything logarithmic is considered, and once the power stands alone the right side turns out to be 2^3, so no logarithm is needed at all. Taking log of 5 · 2^x + 3 = 43 as a first move is legal and leads nowhere useful, because log(5 · 2^x + 3) cannot be expanded: there is no law for the logarithm of a sum.
In equation (3) the answer is ln 30 / ln 7. The decimal is a rounding of that quotient to six places, and it is not the answer — if the value feeds into anything else, carry the quotient and round once, at the end.
For a faded solve, use 4·3^(2x − 1) − 1 = 35. The isolation is supplied: 3^(2x − 1) = 9. Decide whether a logarithm is still needed, finish the exact solution, and substitute it into the original equation including the coefficient and subtraction. Name the feature of 9 that makes your chosen method the shorter one.
Given ten equations you should be able to sort them without solving any: these have both sides as powers of one base and give exact answers, these need a logarithm, and these have no solution because the isolated exponential is not positive. The sorting is the skill. The algebra after it is one or two lines.
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