Odpowiedź:
9.
f ( x) = ( [tex]\frac{1}{5} )^{x - 8} + 2[/tex]
g ( x) = [tex]log_a x[/tex] gdzie a ∈ [tex]R_+[/tex] \ { 1 } x > 0
oraz y = 3
więc
( [tex]\frac{1}{5})^{ x - 8} + 2 = 3[/tex]
( [tex]\frac{1}{5} )^{x - 8} = 3 - 2 = 1[/tex] więc x - 8 = 0
x = 8
P = ( 8, 3)
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a ) 3 = [tex]log_a 8[/tex] ⇔ a = 2 bo 2³ = 8
Odp. g ( x) = [tex]log_2 x[/tex]
=================
b) g( 3) + g( 2) - [tex]2^{g(x)}[/tex] = [tex]log_2 3 + log_2 5 - 2^{log_2 2} = log_2 ( 3*5) - 2^1 = log_2 15- 2 = log_2 15 - log_2 4 = log_2 \frac{15}{4} = log_2 3,75[/tex]
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Verified answer
Odpowiedź:
9.
f ( x) = ( [tex]\frac{1}{5} )^{x - 8} + 2[/tex]
g ( x) = [tex]log_a x[/tex] gdzie a ∈ [tex]R_+[/tex] \ { 1 } x > 0
oraz y = 3
więc
( [tex]\frac{1}{5})^{ x - 8} + 2 = 3[/tex]
( [tex]\frac{1}{5} )^{x - 8} = 3 - 2 = 1[/tex] więc x - 8 = 0
x = 8
P = ( 8, 3)
------------------
a ) 3 = [tex]log_a 8[/tex] ⇔ a = 2 bo 2³ = 8
Odp. g ( x) = [tex]log_2 x[/tex]
=================
b) g( 3) + g( 2) - [tex]2^{g(x)}[/tex] = [tex]log_2 3 + log_2 5 - 2^{log_2 2} = log_2 ( 3*5) - 2^1 = log_2 15- 2 = log_2 15 - log_2 4 = log_2 \frac{15}{4} = log_2 3,75[/tex]
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Szczegółowe wyjaśnienie: