Answer:
1
Explanation:
⅓Log₁₀(125/8) – 2Log₁₀(2/5) + Log₁₀(80/125)
The above expression can be simplified as follow:
⅓Log(125/8) – 2Log(2/5) + Log(80/125)
Recall:
a LogM = Log Mᵃ
Therefore,
⅓Log(125/8) – 2Log(2/5) + Log(80/125)
= Log(125/8)¹/³ – Log(2/5)² + Log(80/125)
= Log(5/2) – Log(4/25) + Log(16/25)
Recall:
Log M – Log N = Log M/N
Log(5/2) – Log(4/25) = Log (5/2 ÷ 4/25)
Log(5/2) – Log(4/25) = Log (5/2 × 25/4)
Log(5/2) – Log(4/25) = Log (125/8)
Therefore,
Log(5/2) – Log(4/25) + Log(16/25)
= Log (125/8) + Log(16/25)
Recall:
Log M + Log N = Log MN
Log (125/8) + Log(16/25)
= Log (125/8 × 16/25)
= Log 10
= Log₁₀10 = 1
Therefore,
⅓Log(125/8) – 2Log(2/5) + Log(80/125)
= 1