rozwiąż równania:
1)
Pierwiastek z 0,2 = (1/5)^(1/2)
zatem mamy
(1/5)^(1/2) = [ 5^2]^x
[ 5 ^(-1)]^(1/2) = 5 ^(2x)
5^( -1/2) = 5^(2x)
-1/2 = 2x / : 2
x = - 1/4 = - 0,25
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2)
4/3 ={ [ 8/27]^x} * 3 / : 3
4/9 = [ 8/27]^x
[2/3]^2 = [ (2/3)^3]^x
[2/3]^2 = [2/3]^(3x)
2 = 3x / : 3
2/3 = x
x = 2/3
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1)
Pierwiastek z 0,2 = (1/5)^(1/2)
zatem mamy
(1/5)^(1/2) = [ 5^2]^x
[ 5 ^(-1)]^(1/2) = 5 ^(2x)
5^( -1/2) = 5^(2x)
-1/2 = 2x / : 2
x = - 1/4 = - 0,25
================
2)
4/3 ={ [ 8/27]^x} * 3 / : 3
4/9 = [ 8/27]^x
[2/3]^2 = [ (2/3)^3]^x
[2/3]^2 = [2/3]^(3x)
2 = 3x / : 3
2/3 = x
x = 2/3
==========