jika 2log1/a=3/2dan 16log b=5maka alog 1/b3???

LogAritMa

2'og 1/a = 3/2
1/a = 2^3/2
a = 2^-3/2

16'log b = 5
b = 16^5
b = 2^20

a'log b
= (2^-3/2)'log 2^20
= 20/(-3/2) . 2'log 2
= -20 . 2/3
= -40/3 ✓

jawab

²log (1/a) = 3/2
²log a⁻¹ = 3/2
- 1 (²log a) = 3/2
²log a = - 3/2


¹⁶log b = 5 → ²^⁴log b = 5
(1/4). ²log b = 5
²log b =  20

ᵃlog 1/b³ = ᵃ log b⁻³ = -3 [ ᵃlog b]
= - 3 [ ²log b/²log  a]
= - 3 [20/(-3/2)]
 = - 3 (-40/3)
=  40