Zadania w załączniku. Proszę o pomoc. :c
4.
a)
(1/5)^x = 125^(-2)
(1/5)^x = (1/125)^2
(1/5)^x = [ (1/5)^3 ]^2
(1/5)^x = (1/5)^6
x = 6
b)
log1/2 x + log1/2 3 = log1/2 5
log1/2 (x * 3) = log1/2 5
3x = 5
x = 5/3
x = 1 i 2/3
c)
log(1-x) 8 = 3 x≠ 1
(1 - x)^3 = 8
(1 - x)^3 = 2^3
1 - x = 2
-x = 2 - 1
-x = 1
x = -1
3.
.... = log1/3 3^3 * 3^(1/2) = log1/3 3^(3 i 1/2) = log1/3 (1/3)^(-3 ,5) = -3,5
.... = log2 (4 * 7 : 14) = log2 (28 : 14) = log2 2 = 1
.... = (3^4)^log3 √2 = 3^log3 (√2)^4 = 3^log3 4 = 4
d)
..... =[ 5^(1/2) * (√5³)³] : 5 = [ 5^(1/2) * (5^(1/3))^3 ] : 5 = [ 5^(1/2) * 5 ] : 5 = √5
2.
y = a^x A = (-4, 16)
x y
16 = a^(-4)
2^4 = (1/a)4
2 = 1/a
2a = 1
a = 1/2
y = (1/2)^x ---- wzór funkcji
1.
(1/3)^π = 3^(-π)
27^(-2/3√3) = 3^(3 * (-2/3√3)) = 3^(-2√3)
(1/9)^(-3/8) = 9^(3/8) = 3^(2 * 3/8) = 3^(3/4)
9^(π/6) = 3^(2 * π/6) = 3^(π/3)
(1/3)^(-2√3) = 3^(2√3)
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4.
a)
(1/5)^x = 125^(-2)
(1/5)^x = (1/125)^2
(1/5)^x = [ (1/5)^3 ]^2
(1/5)^x = (1/5)^6
x = 6
b)
log1/2 x + log1/2 3 = log1/2 5
log1/2 (x * 3) = log1/2 5
3x = 5
x = 5/3
x = 1 i 2/3
c)
log(1-x) 8 = 3 x≠ 1
(1 - x)^3 = 8
(1 - x)^3 = 2^3
1 - x = 2
-x = 2 - 1
-x = 1
x = -1
3.
a)
.... = log1/3 3^3 * 3^(1/2) = log1/3 3^(3 i 1/2) = log1/3 (1/3)^(-3 ,5) = -3,5
b)
.... = log2 (4 * 7 : 14) = log2 (28 : 14) = log2 2 = 1
c)
.... = (3^4)^log3 √2 = 3^log3 (√2)^4 = 3^log3 4 = 4
d)
..... =[ 5^(1/2) * (√5³)³] : 5 = [ 5^(1/2) * (5^(1/3))^3 ] : 5 = [ 5^(1/2) * 5 ] : 5 = √5
2.
y = a^x A = (-4, 16)
x y
16 = a^(-4)
2^4 = (1/a)4
2 = 1/a
2a = 1
a = 1/2
y = (1/2)^x ---- wzór funkcji
1.
(1/3)^π = 3^(-π)
27^(-2/3√3) = 3^(3 * (-2/3√3)) = 3^(-2√3)
(1/9)^(-3/8) = 9^(3/8) = 3^(2 * 3/8) = 3^(3/4)
9^(π/6) = 3^(2 * π/6) = 3^(π/3)
(1/3)^(-2√3) = 3^(2√3)