The companion move to the "same base on both sides, equate exponents" trick is knowing when it doesn't apply. If you get in the habit of equating exponents the moment you see an exponential equation, you will eventually write 2^x = 3^y \Rightarrow x = y, which is wrong for every pair of values except x = y = 0. The common-base move has a prerequisite — that a common base exists — and when the prerequisite fails, you need a different tool.
That tool is logarithms.
The trigger
If both sides of an exponential equation are powers of different primes (or of bases that are not powers of a common smaller base), take logs on both sides.
The failure mode is not subtle: 2 and 3 are both prime, so neither is a power of the other, and there is no smaller base that both are powers of. You cannot rewrite 3 as 2^k for any rational k. The common-base trick has no handle to pull.
The standard example
2^x = 3.
This has a solution — there is some real x that works — but you cannot find it using integer arithmetic. Take \log (any base) of both sides:
The right-hand side is a number — specifically \log_2 3 \approx 1.585. That is the answer.
The crucial step is the middle one: the power rule of logarithms, \log(a^x) = x \log a, pulls the exponent x down from the top and turns it into a coefficient. Once the unknown is a plain coefficient, you divide it out and you have your answer.
Why logarithms handle what common bases can't
The common-base move works because the function a^x is one-to-one: equal outputs force equal inputs. But this only helps when both sides are already written with the same a. If the bases differ, the function itself differs on each side, and you cannot just read off "exponent = exponent."
Logarithms are the inverse of exponentiation, and they work for every positive base. Taking a \log converts "unknown in the exponent" into "unknown times a known \log" — which is a linear equation. Every exponential equation, no matter how the bases are arranged, becomes a linear equation after one well-placed \log.
The general pattern
For equations of the form a^{P(x)} = b^{Q(x)} where a and b have no common base:
- Take \log (any base — natural, base 10, base 2) of both sides.
- Use the power rule to pull the exponents down: P(x) \log a = Q(x) \log b.
- You now have a linear (or quadratic) equation in x. Solve normally.
Once you are comfortable with this, you can use \log_2 or \log_3 or \log_{10} — whichever produces the cleanest constants. For 2^x = 3, taking \log_2 gives x = \log_2 3 directly; taking \log_{10} gives x = \log 3 / \log 2 — same number, different packaging.
A worked JEE Main example
Solve for x:
There is no common base — 5 and 3 are both prime. Take \log of both sides:
Distribute:
Collect x terms:
Solve:
The final form uses the quotient law of logs: 2 \log 3 - \log 5 = \log 9 - \log 5 = \log(9/5). If the problem asked for a decimal, you would compute \log 5 \approx 0.699 and \log(9/5) \approx 0.255, giving x \approx 2.74. If the problem asked for an exact answer, \log_{9/5} 5 is the clean form.
Why collecting x terms like this is the point: after the \log step, x appears in two places, each multiplied by a known constant. This is a linear equation in x, and linear equations are solved by the usual "get x on one side, divide by the coefficient" recipe. The exponential character of the original problem has been absorbed into the \log constants.
The cross-check: is there really no common base?
Before reaching for logs, spend three seconds double-checking that no common base exists. The common-base path is always faster when it works. Ask:
- Are both bases the same? (2^x = 2^7 — yes, common base 2.)
- Is one base a power of the other? (4^x = 8^{x+1} — both powers of 2.)
- Are both bases powers of a smaller integer? (9^x = 27^{x-1} — both powers of 3.)
If any of these is true, use the common-base trick. Only if all three fail should you pivot to logs.
The mixed case: both moves can work
Sometimes the same problem yields to either approach, and it is worth noticing which is cleaner.
4^{x} = 16.
Common-base way: 4 = 4 and 16 = 4^2, so 4^x = 4^2 \Rightarrow x = 2. One line.
Log way: x \log 4 = \log 16 = 2 \log 4, so x = 2. Two lines, same answer.
When the common-base path exists, it is always cleaner. Use it. The log approach is the fallback, not the default.
The small surprise: one constant left
After the log step, the answer often contains \log symbols that do not simplify to nice numbers. \log_2 3 is an irrational number with no closed-form value; \log(9/5) is similar. On JEE, the mark scheme usually accepts the answer in \log form — you do not need to convert to a decimal unless the problem explicitly asks for one. x = \log_2 3 is a complete answer.
This is different from Board problems, where the \log values are often chosen to be neat (\log 10 = 1, \log 2 = 0.301 given) and the question expects a decimal at the end. Read the problem to see what form of answer is expected.
The reflex, in one line
When an exponential equation has different bases that cannot be combined into a common base, take logarithms on both sides and use the power rule \log(a^x) = x \log a to reduce the problem to a linear equation in x. The common-base move is the fast path; logs are the universal fallback.
Related: Exponents and Powers · Same Base Both Sides: Equate Exponents · Rewrite the Base as a Power · What Does a Fractional Exponent Actually Mean?