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Every exponent is a signed contribution.
Worked example · real topic question
(3−6 × 3−5) ÷ 3−9
(83 × 8−4) ÷ 8−2
28 × (2−5 ÷ 26)
Load signs
Switch quotient
Find total
Clear negative
Exponent Switchyard
Route each exponent into one ledger.
Keep the base 3. Products enter as written; quotient exponents pass through the minus gate.
×product lane
−6as written
+
−5as written
× same base → add exponents
÷quotient lane
−9denominator
subtract
→
+9−(−9)
÷ same base → subtract the denominator exponent
signed exponent ledger
−6
+
−5
+
9
=
−2
3−6 + (−5) − (−9) = 3−2
negative total → reciprocal
3−2→132= 1/9
Same switchyard. New exponents.
The denominator has exponent −2, so the quotient gate contributes +2.
(83 × 8−4) ÷ 8−2
+3product
−4product
+2subtract −2
+3 + (−4) − (−2)
3 + (−4) + 2 = 1
81 = 8
Your turn: route all three exponents.
Find the signed exponent total, then clear the negative exponent.
28 × (2−5 ÷ 26)
Keep base 2. The two product exponents enter as written. The denominator exponent is subtracted.
Enter the net exponent and the denominator of the reciprocal.
Use your whiteboard.Write 8 + (−5) − 6. Find the signed total, then rewrite the power with a positive exponent. Submit when your final answer is ready.
Work submitted — the switchyard solution is unlocked.
Read the same three routes.
Every sign in the ledger comes from its lane.
28 × (2−5 ÷ 26)
+8product
−5product
−6quotient
add products · subtract quotient
8 + (−5) − 6 = −3
2−3
negative total → reciprocal
2−3 = 1 / 23 = 1/8
a−n
Switchyard cleared.
You can trace every sign instead of guessing it.
1Keep the common base
2Add product exponents
3Subtract quotient exponents
4Reciprocal if total is negative
The minus in a quotient belongs to the whole denominator exponent. That is why subtracting a negative contribution becomes addition.