4⋅5−6t=2000
which of the following is the solution of the equation?
choose 1
choose 1

(choice a)
a
t=\dfrac{-\log_{2000}(20)}{6}t=
6
−log
2000

(20)

t, equals, start fraction, minus, log, start base, 2000, end base, left parenthesis, 20, right parenthesis, divided by, 6, end fraction

(choice b)
b
t=\dfrac{-\log_{20}(2000)}{6}t=
6
−log
20

(2000)

t, equals, start fraction, minus, log, start base, 20, end base, left parenthesis, 2000, right parenthesis, divided by, 6, end fraction

(choice c)
c
t=\dfrac{-\log_{500}(5)}{6}t=
6
−log
500

(5)

t, equals, start fraction, minus, log, start base, 500, end base, left parenthesis, 5, right parenthesis, divided by, 6, end fraction

(choice d, checked)
d
t=\dfrac{-\log_5(500)}{6}t=
6
−log
5

(500)