Question

vreau si eu E3(d) E4(b,d)

Answer 1

3/d
㏒(1/x)  sa convenim sa scriem asa baza (1/x) si in paranteza [ ]restul
log(1/x) [(7+x)/10]=log(1/x)[2/(x+6)]
conditia de existenta
1/x>0 si 1/x≠1 deci x∈(1,+∞)
(7+x)/10>0⇒x>-7
din cele 2 ⇒x∈(1,+∞)
la a doua 
1/x>0 si 1/x≠1 deci x∈(1,+∞)
x+6≠0,x≠-6,⇒x∈(1,+∞)
(7+x)/10=2/(x+6)
(7+x)(x+6)=20
x²+13x+42-20=0
x²+13x+22=0
Δ=b²-4ac=169-88=81
x1=(-13-9)/2=-22/2=-11
x2=(-13+9)/2=-4/2=-2

b)
notam lgx=y si avem
1/6y²-5/12y-1=0
aducem la acelasi numitor
2y²-5y-12=0
Δ=25+96=121
x1=3/2
x2=4
revenind la notatie
lgx=4⇒lgx=lg10^4⇒x=10^4
lgx=3/2 ⇒lgx=lg 10^3/2⇒x=10^3/2 sau ca la tine in carte x=10^(-2/3)