This is a drill problem. Two powers of x, multiplied together — so the product rule for exponents applies: same base, add the exponents. The only twist is that the exponents are fractions, and fractions cannot be added until they share a denominator. Every student who gets this problem wrong gets it wrong at the fraction step, not at the exponent step. You will land on x^{5/6}, and the whole derivation is three lines long.
Step 1 — recognise the rule
The product law of exponents says:
The law makes no distinction between integer, fractional, or negative exponents. It works for x^3 \cdot x^4 = x^7, for x^{2/3} \cdot x^{1/6}, and for x^{-1/2} \cdot x^{3/4} alike. The only requirement is that the two powers share the same base. Here both bases are x, so you are cleared to add the exponents. Do not take logs, do not convert to radicals — the product rule handles this in one move, as long as you can add two fractions correctly.
Step 2 — add the fractional exponents, common denominator
The exponents are \tfrac{2}{3} and \tfrac{1}{6}. You need \tfrac{2}{3} + \tfrac{1}{6}.
Different denominators. You cannot just stack numerators. Apply the common-denominator routine: find the LCM of the two denominators, rewrite each fraction in terms of that LCM, then add.
LCM of 3 and 6: since 6 is a multiple of 3, the LCM is simply 6. No new number to invent.
Rewrite \tfrac{2}{3} with denominator 6. Multiply top and bottom by 2:
\tfrac{1}{6} already has denominator 6, so it stays as \tfrac{1}{6}.
Now the two fractions are in matching units — sixths — and the numerators add directly:
The fraction \tfrac{5}{6} is already in lowest terms (gcd of 5 and 6 is 1), so no further reduction.
Step 3 — write the answer
Plug the sum back into the product rule:
That is the simplified form. No calculator needed, no decimal approximation, just common-denominator arithmetic sitting on top of one exponent law.
Common mistakes on this class of problem
Three mistakes come up again and again on fractional-exponent products. All three are silent — the wrong answer looks plausible, and the grader has to hunt for where it went off.
Mistake 1: adding numerators and denominators separately. Writing \tfrac{2}{3} + \tfrac{1}{6} = \tfrac{3}{9}. This is the single most famous fraction mistake in school arithmetic. Fraction addition is not component-wise. You cannot add the 2 and 1 on top, then add the 3 and 6 on the bottom, and call that the sum. It ignores the whole point of a denominator — the size of the piece being counted. Halves and sixths are different-sized pieces, and you cannot count them together until you convert them to the same size.
Mistake 2: multiplying the fractions instead of adding them. Writing \tfrac{2}{3} \cdot \tfrac{1}{6} = \tfrac{2}{18} = \tfrac{1}{9}, giving x^{1/9} as the "answer." This confuses two different exponent laws. The product rule x^a \cdot x^b = x^{a + b} adds the exponents. The power-of-a-power rule (x^a)^b = x^{ab} multiplies them. Your problem is two powers multiplied together, not a power raised to a power, so you add — you do not multiply.
Mistake 3: dropping the common-denominator step and guessing a decimal. Computing \tfrac{2}{3} \approx 0.667 and \tfrac{1}{6} \approx 0.167, so 0.833, and writing the answer as x^{0.83}. The arithmetic is fine, but the form is wrong — a textbook answer expects the clean fractional form x^{5/6}. Decimal exponents hide the structure and lose marks. Keep fractional exponents as fractions until the final line.
Sanity-check with a concrete value
A concrete number check takes ten seconds and catches every sign and arithmetic error. Pick x = 64, because 64 = 2^6, so all the sixth roots come out as clean integers.
Compute each factor first:
The cube root of 64 is 4 because 4^3 = 64. Squaring gives 16.
The sixth root of 64 is 2 because 2^6 = 64.
Now multiply the two factors:
Compare against your simplified form x^{5/6} at x = 64:
Both sides give 32. The derivation is correct.
The check works cleanly because 64 is a perfect sixth power — every fractional exponent hits an integer root. When drilling problems of this kind, pick values like 64 = 2^6 or 729 = 3^6; they let you verify without a calculator.
Two more worked drills
Same mechanic, different numbers.
Drill 1. Simplify x^{3/4} \cdot x^{1/2}.
Product rule: add the exponents. LCM of 4 and 2 is 4. Rewrite \tfrac{1}{2} = \tfrac{2}{4}. Sum: \tfrac{3}{4} + \tfrac{2}{4} = \tfrac{5}{4}. Answer:
Drill 2. Simplify x^{1/3} \cdot x^{-1/6}.
Product rule: add the exponents, even when one is negative. LCM of 3 and 6 is 6. Rewrite \tfrac{1}{3} = \tfrac{2}{6}. Sum: \tfrac{2}{6} + \left(-\tfrac{1}{6}\right) = \tfrac{2 - 1}{6} = \tfrac{1}{6}. Answer:
Both drills run through the same five-step routine — and that routine is worth writing out explicitly, because it applies to every problem in this family.
Generalising — the procedure
Given any product x^a \cdot x^b where a and b are fractions (possibly negative), the procedure is:
- Check the bases are the same. If one factor is a power of x and the other is a power of y, the product rule does not apply — skip to the next section.
- Identify the two exponents. Write them out as fractions, keeping signs visible.
- Find the common denominator of the exponent fractions. Usually the LCM of the two denominators, sometimes their product if they share no factor.
- Add the rewritten fractions. Numerators add directly once the denominators match; keep track of negative signs.
- Reduce if possible. If numerator and denominator share a factor, cancel it. The cleanest exam answer is always in lowest terms.
Five steps, done in one line once you are fluent. The routine never changes — the only variation is the specific fractions.
When the bases are not the same — you can't apply the product rule
If the problem is x^{2/3} \cdot y^{1/6}, the bases are different. The product rule requires the same base, so it does not apply here. The expression stays as x^{2/3} \cdot y^{1/6}, already in simplest form. There is no way to combine the exponents, because adding \tfrac{2}{3} and \tfrac{1}{6} would require x and y to be the same variable — which they are not.
Students sometimes try to "average" the exponents, or write (xy)^{5/6}. Wrong — (xy)^{5/6} expands to x^{5/6} \cdot y^{5/6}, which has both factors to the \tfrac{5}{6}, not x to the \tfrac{2}{3} and y to the \tfrac{1}{6}. Different expressions, different values. Different bases, no product-rule shortcut.
What if the exponents are negative
Negative fractional exponents do not change the procedure — they only change the arithmetic. Take x^{-2/3} \cdot x^{1/6}.
Step 1, 2: same base x, exponents -\tfrac{2}{3} and \tfrac{1}{6}.
Step 3: LCM of 3 and 6 is 6. Rewrite -\tfrac{2}{3} = -\tfrac{4}{6}.
Step 4: add. -\tfrac{4}{6} + \tfrac{1}{6} = \tfrac{-4 + 1}{6} = -\tfrac{3}{6}.
Step 5: reduce. -\tfrac{3}{6} = -\tfrac{1}{2}.
Answer: x^{-1/2}. By the negative-exponent rule, x^{-1/2} = \tfrac{1}{x^{1/2}} = \tfrac{1}{\sqrt{x}}, if the problem wants positive exponents or radical form.
The negative sign rides through the addition as ordinary signed-fraction arithmetic. If you can handle -\tfrac{4}{6} + \tfrac{1}{6} without panic, you can handle negative fractional exponents without a new rule.
Close
The product rule for same-base powers is one line: add the exponents. When the exponents are fractions, the only extra work is standard fraction addition — find the common denominator, rewrite, add the numerators, reduce. Skip the common denominator and you will fall into one of the three famous fraction traps. Do the LCM step explicitly, keep the arithmetic clean, and x^{2/3} \cdot x^{1/6} = x^{5/6} drops out in three lines. Same base, add the exponents, mind the fractions. That is the whole problem.