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Kim solved the equation below by graphing a system of equations.What is the approximate solution to the equation?
A
0.6B
0.9C
1.4D
1.62
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A teacher used the change of base formula to determine whether the equation below is correct.Which statement explains whether the equation is correct?
A
The equation is correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = log (2 times 10) times log (4 times 8) times log (10 times 4). = log (20) times log (32) times log (40). =3
B
The equation is correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = StartFraction log 10 Over log 2 EndFraction times StartFraction log 8 Over log 4 EndFraction times StartFraction log 4 Over log 10 EndFraction. = StartFraction log 8 Over log 2 EndFraction. = 3
C
The equation is correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = StartFraction log 10 Over log 2 EndFraction times StartFraction log 8 Over log 4 EndFraction times StartFraction log 4 Over log 10 EndFraction. = StartFraction log 8 Over log 2 EndFraction. = 4.
D
The equation is not correct since (log Subscript 2 Baseline 10) (log Subscript 4 Baseline 8) (log Subscript 10 Baseline 4) = log StartFraction 10 Over 2 EndFraction times log eight-fourths times log four-tenths. = log 5 times log 2 times log 0.4. = negative 0.08
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