⏳Exponential & Logarithmic Equations

Doctor Kelvin keeps two tools on the bench at Exponential Frost, and the first thing to learn is which one to pick up.

Tool one: match the bases. If both sides of the equation can be written as powers of the *same* base, you are already finished, because a power function never gives two different exponents the same value. So

2^x = 32 becomes 2^x = 2⁵, giving x = 5

and once the bases agree you simply set the exponents equal. This works far more often than students expect, because the numbers in a question are usually chosen kindly:

9^x = 27 -> (3²)^x = 3³ -> 3^2x = 3³ -> 2x = 3, x = 3/2 (1/4)^x = 8 -> 2^-2x = 2³ -> -2x = 3, x = -3/2 5^(2x - 1) = 125 -> 2x - 1 = 3, x = 2

Notice what rewriting a base does: 9^x is (3²)^x = 3^2x, so the inner power multiplies the whole exponent, brackets and all. That is the single step people drop.

Tool two: take a logarithm of both sides. When the two sides cannot be made into powers of one base — 3^x = 20 is the classic — no amount of staring helps. Take a logarithm instead. Any base works; ln is the usual choice because every calculator has it.

3^x = 20 ln(3^x) = ln(20) x * ln(3) = ln(20) (the power law brings x down) x = ln(20) / ln(3) = 2.727

That middle line is the whole trick of the topic: the power law turns an exponent into a coefficient, and a coefficient can be divided away.

Two warnings, both about the last line. It is a quotient of logarithms, ln(20) / ln(3), not ln(20) - ln(3) — that would be ln(20/3), a completely different number. And if anything is multiplying the power, isolate it first: 5 * 2^x = 80 becomes 2^x = 16 before you touch the exponent, and then it turns out you did not need a logarithm at all, since x = 4.

Now turn the equation around. When the unknown is *inside* the logarithm, the move is to rewrite the equation in exponential form — the same statement, said the other way.

log_b(A) = c means exactly b^c = A

Read that as "the logarithm IS the exponent". So log_2(x) = 5 says "2 to the power 5 gives x", and x = 32. The commonest slip is to multiply instead: 2 * 5 = 10 is not it, and neither is 5². The base stays the base.

Once you can flip the statement, everything else is ordinary equation solving.

log_2(x - 3) = 4 -> x - 3 = 2⁴ = 16 -> x = 19 ln(x) = 2 -> x = e² = 7.389 log(2x + 1) = 2 -> 2x + 1 = 100 -> x = 99/2

Two conventions to keep straight. A logarithm written with no base at all is base 10 — log(1000) = 3. And ln means base e, so ln(x) = 2 gives e², never 10².

Isolate the logarithm before you rewrite. If there is a coefficient or a constant hanging around, deal with it first, exactly as you would for any other equation:

3 * log_2(x - 5) + 4 = 16 3 * log_2(x - 5) = 12 (take the 4 off) log_2(x - 5) = 4 (divide by 3) x - 5 = 2⁴ = 16 (now rewrite) x = 21

Rewriting too early is what turns a small question into a monster: raising a base to the power 12 instead of 4 is not a rounding difference, it is a different planet.

Some equations arrive with two or three logarithms in them. They cannot be rewritten one at a time — you have to condense them into a single logarithm first, using the laws from the last lesson:

• ◆log_b(A) + log_b(B) = log_b(A * B) — the arguments multiply.
• ◆log_b(A) - log_b(B) = log_b(A / B) — the arguments divide.
• ◆n * log_b(A) = log_b(Aⁿ) — a coefficient becomes a power.

Then either rewrite in exponential form, or, if both sides are single logarithms with the same base, just match the arguments (log_b(A) = log_b(B) gives A = B, because a logarithm is one-to-one).

log_2(x) + log_2(x - 2) = 3 log_2(x(x - 2)) = 3 (product law: multiply, do NOT add) x(x - 2) = 2³ = 8 (exponential form) x² - 2x - 8 = 0 (x - 4)(x + 2) = 0 so x = 4 or x = -2

Here is where the topic earns its reputation. x = -2 is not a solution. Put it back into the original equation and it asks for log_2(-2), and no power of 2 is ever negative, so that logarithm has no value at all. The candidate is extraneous — it exists only because condensing two logarithms into one quietly widened the domain. The answer is x = 4, alone.

So the last line of every one of these questions is a domain check: every argument of every logarithm, tested on the original equation, must come out strictly positive. Zero is not allowed either — log_b(0) has no value.

That check is also a question in its own right. log_2(x - 5) is defined exactly when x - 5 > 0, that is x > 5 — strictly greater, because at x = 5 the argument is 0. And an equation can lose *every* candidate: log_2(x) - log_2(x - 4) = -1 condenses to x/(x - 4) = 1/2, which gives x = -4, which the domain rejects. That equation has no solution, and saying so is the correct answer.

The reason anyone learned any of this: the unknown that sits in an exponent is almost always time.

Time to reach an amount. A herd of 500 grows 4% a year. When does it reach 800?

500 * 1.04^t = 800 1.04^t = 1.6 (divide first — always) t * ln(1.04) = ln(1.6) t = ln(1.6) / ln(1.04) = 12.0 years

If the question asks "after how many whole years does it first pass 800", round that answer up, never to the nearest: the herd is still short of the target for the whole of the year it crosses in.

Doubling time and half-life. With continuous growth A(t) = A0 * e^kt, the starting amount divides out and never matters:

A0 * e^kt = 2 * A0 -> e^kt = 2 -> t = ln(2) / k

so a colony growing continuously at 5% per hour doubles in ln(2) / 0.05 = 13.9 hours. Half-life is the same sum with 0.5 in place of 2, and if you are only given two readings, find k first: 200 grams falling to 150 grams in 30 years gives k = ln(150/200) / 30, and then the half-life is ln(0.5) / k = 72.3 years.

The named scales. Three of the world's measuring scales are logarithms, which is why each of them hides a power of ten.

• ◆Richter, M = log(I / I0). Rearranged, I = I0 * 10^M, so two tremors three whole magnitudes apart differ in intensity by 10³ = 1000 times, not by 3.
• ◆Decibels, L = 10 * log(I / I0), so I = I0 * 10^L/10 and a sound 30 dB louder is 10³ = 1000 times as intense.
• ◆pH, pH = -log(H), where H is the hydrogen ion concentration. A concentration of 10^-4 moles per litre gives pH 4 — the minus sign in the formula is what makes pH a friendly positive number.

Cooling and dating work the same way: set the model equal to the value you want, peel off anything added on (the room temperature in Newton's cooling law lives *outside* the exponential term, so subtract it before dividing), take ln, and divide by the rate.

Doctor Kelvin marks by value, but the form matters, and every prompt says which form it wants.

Exact answers. When the bases match, or a logarithm rewrites to a whole power, the answer is exact — type it as an integer or a fraction in lowest terms: 5, -6, 3/2, -3/2, 99/2. Do not turn 3/2 into 1.5 when the prompt asked for a fraction, and never leave 6/4 unsimplified.

Calculator answers. When the answer runs on forever, the prompt says "rounded to 3 decimal places": type 2.727, not ln(20)/ln(3) and not 2.73. Keep the full decimal in your calculator until the very last line — rounding ln(20) to 3.0 before dividing will move the answer far more than you expect.

Times. A time is usually "rounded to 1 decimal place" — 13.9, 72.3 — and the unit is written in the prompt, so you can type just the number. When the prompt asks for whole years, give the whole number you get by rounding up: 8, not 7 and not 7.5.

Questions with options. Anything that is not a single number arrives as a multiple choice, because a general base like log_2 cannot be typed into the answer box and neither can an inequality. That covers "which candidate must be rejected, and why", "how many solutions does this have", and "for which x is this defined" (answers like x > 5). Pick the option; you are not expected to type the inequality.

One last habit. Before you send an answer to a condensed equation, put it back into the equation you were given — not into the quadratic you made — and check that every logarithm is being handed a positive number. That thirty-second check is the difference between the right answer and a right-looking one.