Algebra: a1) $2+x=7 rarr x=7-2$ a2) $2x=7 rarr x=7/2$ a3) $x-2=7 rarr x=9$ a4) $x/2=7 rarr x=14$ | regole: r1) $a^x a^y=a^(x+y)$ (prodotto stessa base) r2) $a/b^n= ab^(-n)$ (potenza negativa) r3) $root(m)(a^n)=a^(n/m)$ (potenza frazionaria) r4) $((a)^x)^y=a^(xy)$ (potenza di potenza) r5) $log a^x= xloga$ regola "del fischio": r6) $loga + log b = logab$ (somma di logaritmi) r7) $loga - log b = log(a/b)$ (differenza di logaritmi) r8) $log_a b = (log_cb)/ (log_ca) $ (cambio di base) | NON REGOLE: $(log a)/(log b)$ =? non si puo' fare nulla: non e' logaritmo del rapporto ma rapporto di logaritmi!! $ log(a+b) $ =? non si puo' fare nulla: non e' somma di logaritmi ma logaritmo della somma $ log(a-b) $ =? non si puo' fare nulla: non e' differenza di logaritmi ma logaritmo della differenza | trucchi: t1) $8rarr2^3,(1/8)rarr 2^(-3)$ trucco 1: cerco di portare tutto alla stessa base t2) $2^(x+3)rarr (2^x)(2^3) rarr (8)(2^x)$ se vedo una somma a esponente "..." smonto t3) $(2)(3^x)-(5)(3^(2x))=0 rarr 2t-5t^2=0$ sostituisco per ottenere una eq. di 2° grado in t $t_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)$ t4) $a^x=a^y rArr x=y$ | |||||||
$ 2^x=4root(3)(2) $ (t1+r3) $ 2^x=2^2 2^(1/3) $ (r1) $ 2^x = 2^(2+1/3)$ (eseguo la somma) $ 2^x=2^(7/3) $ (t4) $ x=7/3$ |
$3^x (3^(1/2))/(3^(x/3))=3^2$ (r2) $(3^x)(3^(1/2))(3^-(x/3))=3^2$ (r1) $3^((6x+3-2x)/(6))=3^2 $ eseguo la somma $3^((4x+3)/6)=3^2$ (t4) $(4x+3)/6=2 $ (a4) $4x+3=12 $ (a1) $4x=9 $ (a2) $x=9/4$ |
$7^(x/3)=5$ $ log(7^(x/3))=log(7^(x/3)) $ $ (x/3)log7=log5 $ $ (x/3)=log5/log7 $ $ x=3log5/log7$ |
$3^x+20=(3^2)^x$ $ 3^x+20=3^(2x) $ $ 3^x+20=(3^x)^2 $ $ t+20=t^2 $ $ t^2-t-20=0 $ $ t_(1,2)=(1+-sqrt(1+80))/(2) $ $ t_(1,2)=(1+-9)/(2) $ $ t_1=10/2=5 $ $t_2=-8/2=-4 $(impossibile perche' $t_2=3^x=-4$ non ha soluzioni $a^x>0 forall x$ ) $ t_1=5 $ $ 3^x=5 $ $ log(3^x)=log(5) $ $ xlog3=log5 $ $ x=log5/log3$ | |||||||
$2^3 2^(2x)-25 2^x +3 = 0 $ $ 8t^2-25t+3=0 $ $....4 |
$(3)(3^x) + 3/(3^x)=10 $ $ (3^x (3)(3^x)+3)/(3^x)=10 $ $ ((3)(3^(2x))+3)/(3^x)=10 $ $ ((3)(3^(2x))+3)=10(3^x) $ $ 3t^2 +3 = 10t $ $ 3t^2 -10t + 3 = 0 $ |
$2/5^x=3/7^x$ $2/3=(5^x/7^x)$ $2/3=(5/7)^x$ $log(2/3)=log(5/7)^x$ $log2 -log3 =xlog(5/7)$ $log2 -log3 =x(log5 - log7)$ $(log 2 - log 3)/(log 5 - log 7) = x$ |
$3^x+3^(x+1)=5^x$ $3^x+(3)3^x=5^x$ $(4)3^x=5^x$ $(4)=5^x/3^x$ $(2^2)=(5/3)^x$ $log(2^2)=log((5/3)^x)$ $2log2=xlog(5/3)$ $(2log2)/(log(5/3))=x$ $(2log2)/(log5-log3)=x$ |
$(3)5^x-12/5^x=5^x$ $((5^x)(3)(5^x)-12)/5^x=5^x$ $(5^x)(3)(5^x)-12=(5^x)(5^x)$ $(5^x)(5^x)3-12=5^(2x)$ $3(5^(2x)-12=5^(2x)$ $3(5^(2x)-5^(2x)=12$ $2(5^(2x))=12$ $5^(2x)=12/2$ $5^(2x)=6$ $log(5^(2x))=log6$ $2xlog5=log6$ $2x=log6/log5$ $x=log6/(2(log5))$ $x=log6/(2log5)$ | ||||||
$2^(sqrt(x))=(root(5)(2))/(root(7)(2))$ $2^(sqrt(x))=2^(1/5)/2^(1/7)$ $2^(sqrt(x))=2^(1/5-1/7)$ $2^(sqrt(x))=2^((7-5)/(35))$ $2^(sqrt(x))=2^((2)/(35))$ $sqrt(x)=2/35$ $(sqrt(x))^2=(2/35)^2$ $x=4/1225$ |
$(7)(2^x)+5/(2^x)=117/4$ $((2^x)(7)(2^x)+5)/(2^x)=117/4$ $(7(2^x)(2^x+5))/(2^x)=117/4$ $(7(2^(2x)+5))/(2^x)=117/4$ $7(2^(2x)+5)=(117/4)(2^x)$ $7(2^(2x))-(117/4)(2^x)+5=0; rarr ; t=2^x; t^2=2^(2x)$ $7t^2-117/4t+5=0$ $28t^2-117t+20=0$ $t_(1,2)=(117+-sqrt(117^2-4*28*20))/(56)$ $t_(1,2)=(117+-sqrt(11449))/(56)$ $t_(1,2)=(117+-107)/(56)$ $t_1=(117+107)/(56)=224/56=4$ $t_2=(117-107)/(56)=10/56=5/28$ $t_1) 2^x=4 $ |
$t_2) 2^x=5/28 $ $ log(2^x)=log(5/28) $ $ xlog_2 2=log_2 5-log_2 28 $
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$(6)(2^x)+1/(2^x)=5$ $6(2^x)(2^x)+1=5(2^x)$ $6t^2-5t+1=0$ $t_(1,2)=(5+-sqrt(25-24))/(12)$ $t_(1,2)=(5+-1))/(12)$ $t_1=6/12=1/2$ $t_2=4/12=1/3$ $t_1) 2^x=1/2$ $2^x=2^(-1)$
$t_2) 2^x=1/3$ $log_2 2^x= log_2 (1/3)$ $x log_2 2 = log_2 1 - log_2 3 $ $x(1)= 0 - log_2 3$ $x=log_2 3$
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$3^(sqrt(x-3))=9^(sqrt(x))$ $3^(sqrt(x-3))=3^(2sqrt(x))$ $log_3 (3^(sqrt(x-3))) = log_3 3^(2sqrt(x))$ $sqrt(x-3)=2sqrt(x)$ $x-3=4x$ (devo imporre x>0 e x-3>0) $-3x-3=0$ $3x+3=0$ $3(x+1)=0$ $x=-1$ ma x era sotto radice quindi va' rifiutata |
$2^x-(1)/(2(root(3)(2)))(3^x-5)=0$ soluzione1: $3^x-5=0$ $3^x=5$ $xlog3=log5$ $x=log5/log2$ soluzione2: $2^x-(1)/(2(root(3)(2)))=0$ $2^x=(1)/(2(root(3)(2)))$ $2^x=1/ ( 2^(1) (2^(1/3) ) $ $2^x=1/2^(1+1/3)$ $2^x=1/2^(4/3)$ $2^x=2^(-4/3)$ $x=-4/3$ |
$3^(sqrt(x+2))=9^(sqrt(x))$ $3^(sqrt(x+2))=3^(2sqrt(x))$ $log_3 (3^(sqrt(x+2))) = log_3 3^(2sqrt(x))$ $sqrt(x+2)=2sqrt(x)$ x>0 e x>-2 ! (affinche' le radici siano positive) $x+2=4x$ $-3x+2=0$ $-3x=-2$ $x=-2/(-3)$ $x=2/3$ |
$log_2(x-1)=3$ $2^3=x-1$ (per definizione di logaritmo) $8=x-1$ $9=x$ $x=9$ | |||||||
$log_(0.6)(2x+1)=log_(0.6)x$ (x>0 ! , 2x+1 >0 !) $2x+1=x$ $x+1=0$ $x=-1$ : impossibile perche' x deve essere >0 |
$log_3 6 + log_3 (x+1)= log_3 5x$ $log_3 (6(x+1))=log_3 5x $ x>0 ! x+1>0 $6(x+1)=5x$ $6x-5x+6=0$ $x+6=0$ $x=-6$ impossibile |