Forumet - Ma C potenser

Ma C potenser

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Spana också in:

Holymacarony:


Korrigering av kursbokens felaktiga lösning:

3*9^x = 1/81
log(3*9^x) = log(1/81)
log(3)+x*log(9) = log(1)-log(81) = -log(81)
x = (-log(81)-log(3))/log(9) = -2,5

Två alternativa lösningar:

3*9^x = 1/81
3¹*(3²)^x = 1/3^4
3¹*3^(2x) = 3^(-4)
3^(1+2x) = 3^(-4)
1+2x = -4
1+2x-1 = -4-1
1-1+2x = -5
0+2x = -5
2x = -5
2x/2 = -5/2
2/2*x = -5/2
1*x = -5/2
x = -5/2
x = -2,5

3*9^x = 1/81
3¹*(3²)^x = 1/3^4
log(3¹*(3²)^x) = log(1/3^4)
log(3¹)+log((3²)^x) = log(1)-log(3^4)
1*log(3)+x*log(3²) = 0-4*log(3)
1*log(3)+2x*log(3) = -4*log(3)
(1+2x)log(3) = -4*log(3)
(1+2x)log(3)/log(3) = -4*log(3)/log(3)
(1+2x)*1 = -4*1
1+2x = -4
1+2x-1 = -4-1
1-1+2x = -5
0+2x = -5
2x = -5
2x/2 = -5/2
2/2*x = -5/2
1*x = -5/2
x = -5/2
x = -2,5