ORIGINAL WORKED EXAMPLE
Apply the idea
Simplify (3⁴×3²)/3³.
- The numerator is 3^(4+2)=3⁶.
- Divide by 3³ to obtain 3^(6−3)=3³.
- Evaluate the remaining small power.
Answer 27.
GMAT · Arithmetic and number properties · TOPIC LESSON
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
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01 / UNDERSTAND & APPLY
For integer exponents, multiplying powers with the same base adds exponents: a^m a^n=a^(m+n). Dividing subtracts exponents when a≠0. Raising a power to a power multiplies exponents. A power distributes over a product, not over addition: (a+b)²=a²+2ab+b².
Parentheses determine the base. For integer exponents, (−2)^4=16, while −2^4=−16. Do not apply integer-power laws without checking real-number domains when exponents are fractional.
ORIGINAL WORKED EXAMPLE
Simplify (3⁴×3²)/3³.
Answer 27.
ORIGINAL PRACTICE · 1
No: 125 versus 35.
The left side is 5³=125. The right side is 8+27=35. Powers do not distribute over sums; the missing cross terms matter.
02 / UNDERSTAND & APPLY
For nonzero a, a^0=1 because a^m/a^m=1. A negative exponent gives a reciprocal: a^(−n)=1/a^n. It does not put a minus sign in front of the answer. Zero to a negative power is undefined, and 0^0 is not assigned a value in these ordinary algebraic rules.
Rewrite reciprocal bases to compare expressions. For example, (1/2)^(−3)=2³. If a denominator contains several factors, a negative exponent on one factor applies only to that factor unless parentheses group the entire denominator.
ORIGINAL WORKED EXAMPLE
Evaluate 2^(−3)+4^0.
Answer 9/8.
ORIGINAL PRACTICE · 1
4/9.
Invert the entire nonzero base and square: (2/3)²=4/9. Neither −9/4 nor 9/4 respects the reciprocal meaning of a negative exponent.
03 / UNDERSTAND & APPLY
For a positive real number a, a^(m/n) means the nth root of a raised to integer power m, with n positive. Choosing whether to root or exponentiate first can keep arithmetic small. Negative exponents still take reciprocals.
For negative real bases, use fractional exponents cautiously: write the exponent in lowest terms and require an odd root denominator for a real root interpretation. Even roots require nonnegative radicands. In no-calculator work, recognize small perfect powers instead of approximating roots.
ORIGINAL WORKED EXAMPLE
Evaluate 81^(3/4).
Answer 27.
ORIGINAL PRACTICE · 1
1/4.
The fifth root of 32 is 2. Squaring gives 4, and the negative exponent takes the reciprocal. The denominator 5 indicates a root, not multiplication by 1/5.
04 / UNDERSTAND & APPLY
The square-root symbol denotes the nonnegative root. For nonnegative a and b, √(ab)=√a√b. Pull out perfect-square factors: √72=√(36×2)=6√2. Terms involving the same remaining radical can be combined like coefficients.
A square root does not distribute over addition: √(9+16)=5, not 7. Also √(x²)=|x| for real x. Replacing it by x silently assumes x≥0, which may exclude allowed negative values.
ORIGINAL WORKED EXAMPLE
Simplify √75−√12+√27.
Answer 6√3.
ORIGINAL PRACTICE · 1
√(x²)=7, which is not x.
x²=49 and the principal square root is 7. In general √(x²)=|x|. The squared equation y²=49 has two roots, but the expression √49 has only the nonnegative value.
05 / UNDERSTAND & APPLY
If the base is greater than 1, larger exponents produce larger positive powers. For a base strictly between 0 and 1, the order reverses. With equal positive exponents, larger positive bases produce larger values. Negative bases require sign and parity analysis before magnitude comparison.
Try rewriting both powers using a common base or exponent. If that fails, use an exact bound rather than a decimal approximation. Squaring preserves order for nonnegative quantities, but not for arbitrary signed quantities.
ORIGINAL WORKED EXAMPLE
Which is larger, 8⁷ or 4¹¹?
Answer 4¹¹, which is twice 8⁷.
ORIGINAL PRACTICE · 1
(1/3)³ < (1/3)² < (1/3)^(−1).
The values are 1/27, 1/9 and 3. Positive powers shrink when the base is between 0 and 1; a negative exponent reciprocates the base. Applying the greater-than-1 rule would reverse the first comparison.
06 / UNDERSTAND & APPLY
Scientific notation writes a nonzero magnitude as a×10^n with 1≤|a|<10 and integer n. The power of ten captures scale. To multiply, multiply coefficients and add exponents; to divide, divide coefficients and subtract exponents. Normalize the coefficient afterward.
To add quantities, first express them with the same power of ten. Adding exponents during addition is invalid. A rough order-of-magnitude check is especially useful when units are large or small.
ORIGINAL WORKED EXAMPLE
Compute (6×10⁵)/(2×10^(−3)) in scientific notation.
Answer 3×10⁸.
ORIGINAL PRACTICE · 1
4.7×10⁶.
Rewrite 7×10⁵ as 0.7×10⁶ and add coefficients. Adding exponents would incorrectly describe multiplication rather than addition.
07 / UNDERSTAND & APPLY
In real-number problems, an even root has a nonnegative radicand and produces a nonnegative principal value. Thus √(x+1)=x−1 requires x≥1 even before squaring. Squaring both sides is a one-way implication unless signs are controlled.
After solving a squared or denominator-cleared equation, substitute every candidate into the original expression. The check must use the original domain and original signs, not the transformed equation, which all candidates already satisfy.
ORIGINAL WORKED EXAMPLE
Solve √(x+6)=x.
Answer x=3 only.
ORIGINAL PRACTICE · 1
x=−5 or x=5.
The equation is |x|=5, so both signed values work. Keeping only x=5 confuses the nonnegative square-root output with the possibly negative input x.
08 / UNDERSTAND & APPLY
In GMAT Problem Solving, a large exponent is usually a reason to seek structure, not to carry out a long multiplication. Factor a shared power, compare signs, or determine a cycle. Retain all restrictions while simplifying.
A shortcut is useful only if it answers the requested quantity. Knowing that two powers are both positive does not tell you which is larger. Knowing a last digit does not establish divisibility by 3. State the exact information each simplification preserves.
ORIGINAL WORKED EXAMPLE
Simplify (5⁶−5⁵)/5⁴.
Answer 20.
ORIGINAL PRACTICE · 1
√50 is greater than 7 and less than 8.
Since 49<50<64, taking nonnegative square roots gives 7<√50<8. This bound proves the comparison without approximating the radical; treating √50 as √49+√1 would give an incorrect value.
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