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) $a^x=a^y rArr x=y$


186
$ 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$



187
$3^x (3^(1/2))/(3^(x/3))=3^2$ (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)

$x=9/4$



188
$7^(x/3)=5$

$ log(7^(x/3))=log(7^(x/3)) $

$ (x/3)log7=log5 $

$ (x/3)=log5/log7 $

$ x=3log5/log7$



189
$3^x+20=(3^2)^x$

$ 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$




190
$2^3 2^(2x)-25 2^x +3 = 0 $

$ 8t^2-25t+3=0 $

$....4



191
$(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 $



192
$2/5^x=3/7^x$

$2/3=(5^x/7^x)$

$2/3=(5/7)^x$

$log(2/3)=log(5/7)^x$

$log2 -log3 =xlog(5/7)$

$log2 -log3 =x(log5 - log7)$

$(log 2 - log 3)/(log 5 - log 7) = x$



193
$3^x+3^(x+1)=5^x$

$3^x+(3)3^x=5^x$

$(4)3^x=5^x$

$(4)=5^x/3^x$

$(2^2)=(5/3)^x$

$log(2^2)=log((5/3)^x)$

$2log2=xlog(5/3)$

$(2log2)/(log(5/3))=x$

$(2log2)/(log5-log3)=x$



194
$(3)5^x-12/5^x=5^x$

$((5^x)(3)(5^x)-12)/5^x=5^x$

$(5^x)(3)(5^x)-12=(5^x)(5^x)$

$(5^x)(5^x)3-12=5^(2x)$

$3(5^(2x)-12=5^(2x)$

$3(5^(2x)-5^(2x)=12$

$2(5^(2x))=12$

$5^(2x)=12/2$

$5^(2x)=6$

$log(5^(2x))=log6$

$2xlog5=log6$

$2x=log6/log5$

$x=log6/(2(log5))$

$x=log6/(2log5)$



195
$2^(sqrt(x))=(root(5)(2))/(root(7)(2))$

$2^(sqrt(x))=2^(1/5)/2^(1/7)$

$2^(sqrt(x))=2^(1/5-1/7)$

$2^(sqrt(x))=2^((7-5)/(35))$

$2^(sqrt(x))=2^((2)/(35))$

$sqrt(x)=2/35$

$(sqrt(x))^2=(2/35)^2$

$x=4/1225$



196
$(7)(2^x)+5/(2^x)=117/4$

$((2^x)(7)(2^x)+5)/(2^x)=117/4$

$(7(2^x)(2^x+5))/(2^x)=117/4$

$(7(2^(2x)+5))/(2^x)=117/4$

$7(2^(2x)+5)=(117/4)(2^x)$

$7(2^(2x))-(117/4)(2^x)+5=0; rarr ; t=2^x; t^2=2^(2x)$

$7t^2-117/4t+5=0$

$28t^2-117t+20=0$

$t_(1,2)=(117+-sqrt(117^2-4*28*20))/(56)$

$t_(1,2)=(117+-sqrt(11449))/(56)$

$t_(1,2)=(117+-107)/(56)$

$t_1=(117+107)/(56)=224/56=4$

$t_2=(117-107)/(56)=10/56=5/28$

$t_1) 2^x=4 $



$x=2$
$t_2) 2^x=5/28 $

$ log(2^x)=log(5/28) $

$ xlog_2 2=log_2 5-log_2 28 $

$x=log_2 5 - log_2 28$


197
$(6)(2^x)+1/(2^x)=5$

$6(2^x)(2^x)+1=5(2^x)$

$6t^2-5t+1=0$

$t_(1,2)=(5+-sqrt(25-24))/(12)$

$t_(1,2)=(5+-1))/(12)$

$t_1=6/12=1/2$

$t_2=4/12=1/3$

$t_1) 2^x=1/2$

$2^x=2^(-1)$

$x=-1$
$t_2) 2^x=1/3$

$log_2 2^x= log_2 (1/3)$

$x log_2 2 = log_2 1 - log_2 3 $

$x(1)= 0 - log_2 3$

$x=log_2 3$

$x=log 3 / log 2$



198
$3^(sqrt(x-3))=9^(sqrt(x))$

$3^(sqrt(x-3))=3^(2sqrt(x))$

$log_3 (3^(sqrt(x-3))) = log_3 3^(2sqrt(x))$

$sqrt(x-3)=2sqrt(x)$

$x-3=4x$ (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



199
$2^x-(1)/(2(root(3)(2)))(3^x-5)=0$

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) ) $

$2^x=1/2^(1+1/3)$

$2^x=1/2^(4/3)$

$2^x=2^(-4/3)$

$x=-4/3$



200
$3^(sqrt(x+2))=9^(sqrt(x))$

$3^(sqrt(x+2))=3^(2sqrt(x))$

$log_3 (3^(sqrt(x+2))) = log_3 3^(2sqrt(x))$

$sqrt(x+2)=2sqrt(x)$ x>0 e x>-2 ! (affinche' le radici siano positive)

$x+2=4x$

$-3x+2=0$

$-3x=-2$

$x=-2/(-3)$

$x=2/3$



201
$log_2(x-1)=3$

$2^3=x-1$ (per definizione di logaritmo)

$8=x-1$

$9=x$

$x=9$



202
$log_(0.6)(2x+1)=log_(0.6)x$ (x>0 ! , 2x+1 >0 !)

$2x+1=x$

$x+1=0$

$x=-1$ : impossibile perche' x deve essere >0


203
$log_3 6 + log_3 (x+1)= log_3 5x$

$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


204
$log_(135)(x(x-1))=log_(135)(5x(x-2))$
$x(x-1)=5x(x-2)$
$x^2-x=5x^2-10x$
$-4x^2+9x=0$
$4x^2-9x=0$
$x(4x-9)=0$
soluzione1: x=0 (impossibile: argomenti dei log devono essere >0)
soluzione2: $4x-9=0 rarr x=9/4 $

205
$log(5(x-2))=logx$
$5x-10=x$
$4x=10$
$x=5/2$ accettabile

206
$log((x-3)(x+1))=log(4x-3)$
$x^2-2x-3=4x-3$
$x^2$-6x=0$
$x(x-6)=0$
soluzione1: x=0 imp.
soluzione2: x=6 accettabile

207
$2logx=-2$
$logx=-1$
$x=10^-1$ per definizione di logaritmo
$x=1/10$

208
$log(x-2)=2$
$10^2=x-2$
$100+2=x$
$x=102$

209
$log(x-2)-log(x-1)=log5$
$log((x-2)/(x-1))=log5$
$(x-2)=5(x-1)$
$x-2-5x+5=0$
$-4x+3=0$
$x=-3/(-4)$
$x=3/4$ non accettabile perche' x-1 < 0 argomento del log

210
$log(x+1)+log(x+2)=log2$
$log((x+1)(x+2))=log2$
$x^2+3x+2=2$
$x^2+3x=0$
$x(x+3)=0$
sol1: x=0 accettabile
sol2: x=-3 imp.

211
$2logx=log(x+3)$
$x^2=x+3$
$x^2-x-3=0$
$x_(1,2)=(1+-sqrt(1+13))/(2)$
$x_1=1-sqrt(13)/2$ non accettabile
$x_2=1+sqrt(13)/2$ accettabile

212
$2log(x-1)=logx$
$(x-1)^2=x$
$x^2-2x+1=x$
$x^2-3x+1=0$
$x_(1,2)=(3+-sqrt(9-4))/(2)$
$x_(1,2)=(3+-sqrt(5))/(2)$
accettabile solo $x=(3+sqrt(5))/(2)$

213
$log_2(x)+log_2(x-1)=1$
$log_2(x(x-1))=log_2(2)$
$x(x-1)=2$
$x^2-x-2=0$
$x_(1,2)=(1+-sqrt(1+8))/(2)$
$x_(1,2)=(1+-3))/(2)$
$x_1=2$ accettabile
$x_1=-1$ non accettabile