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marius9505
Pe: 2011-10-29T07:12:19+03:00 In: In:

Probleme logaritmi

1

    \[\begin{array}{l} {\log _{72}}48 = a\\ {\log _6}24 = b\\ \end{array}\]

aratati ca a(b + 3) – 3b + 1 = 0

2) a = {\log _{\sin x}}\cos x\\ unde x apartine (0, pi/2)- {pi/4} exprimati in functie de a log in baza tg(x) din sin2x/2

3)

    \[\begin{array}{l} a = {\log _{\sqrt 2 - 1}}(\sqrt 3 - 1) + {\log _{\sqrt 2 + 1}}(\sqrt 3 + 1)\\ b = {\log _4}({\log _3}({\log _{\frac{1}{2}}}\frac{1}{{512}}) \end{array}\]

verificati daca a) a^2+4b^2=1
b)a>=b
c)a-b=2
d)a=b

4. Calculati sumele
a)

    \[{\log _2} + {\log _2}{a^2} + ... + {\log _2}{a^n}\]

b)

    \[\log a + ... + n\log {a^n}\]

c)

    \[\lg \sqrt[2]{a} + \lg \sqrt[4]{a} + ... + \lg \sqrt[{{2^n}}]{a}\]

5)Demonstrati
a)

    \[{\left( {\frac{a}{b}} \right)^{\lg a}} + {\left( {\frac{c}{a}} \right)^{\lg b}} + {\left( {\frac{a}{b}} \right)^{\lg c}} = 1\]

b)

    \[{\left( {\frac{{ac}}{b}} \right)^{\lg \frac{c}{a}}} + {\left( {\frac{{bc}}{a}} \right)^{\lg \frac{b}{c}}} + {\left( {\frac{{ab}}{c}} \right)^{\lg \frac{a}{b}}} = 1\]

c) daca a,b>0 atunci a^2+4b^2=6ab <=>

    \[\lg (a + 2b) = \frac{1}{2}(\lg 1 + \lg b + 1)\]

d)daca x,y>0 si x^2+y^2=12xy <=>

    \[\lg (x + 2y) - 2\lg 2 = \frac{1}{2}(\lg x + \lg y)\]

6)demonstrati ca
a)

    \[\frac{{{{\log }_x}y}}{{x + y}} + \frac{{{{\log }_y}z}}{{y + z}} + \frac{{{{\log }_z}x}}{{z + x}} \ge \frac{9}{{2(x + y + z)}}\]

, unde x,y,z apartin (0,1) sau (1, + infinit)

b)

    \[6 < \frac{1}{{{{\log }_7}14 - {{\log }_{14}}28}} < 12\]

7) aratati ca numerele

    \[\lg 2,{\log _3}5\]

sunt irationale

8)Fie a,b,c apartin (0, + infinit) – {1}, b diferit de radical din c . Sa se arate ca a,b,c sunt in prog. geometrica <=>

    \[{\log _a}\frac{a}{{{b^2}}} = {\log _{\frac{c}{b}}}a\]

8 )
a)sa se arate ca daca

    \[\begin{array}{l} x = {\log _a}bc\\ y = {\log _b}ac\\ z = {\log _c}ab \end{array}\]

x+y+z+2=xyz
b) fie a,b,c apartine R*+ .Aratati ca daca

    \[\begin{array}{l} {2^a} = 35\\ {5^b} = 14\\ {7^c} = 10 \end{array}\]

abc-(a+b+c)= 2

9) fie a,b,c apartin( o,1) sau (1, + infinit) . Sa se arate ca

    \[{\log _a}x,{\log _b}x,{\log _c}x\]

sunt in prog. geometrica daca si numai daca

    \[{\log _a}b = {\log _b}c\]

10)
aratati ca

    \[\frac{1}{{{{\log }_2}1 + {{\log }_2}2 + ... + {{\log }_2}n}} + \frac{1}{{{{\log }_3}1 + {{\log }_3}2 + ... + {{\log }_3}n}} + ... + \frac{1}{{{{\log }_n}1 + {{\log }_n}2 + ... + {{\log }_n}n}} = 1\]

cam multe exercitii:( dar as fi recunoscator daca mi le-ati putea face pana duminica diseara
am facut cateva greseli(prima data cand fac in mathtype unde nu intelegeti
spunetimi [tex]

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7 raspunsuri

  1. red12dog34
    veteran (III)

    marius9505 wrote: 1

        \[\begin{array}{l} {\log _{72}}48 = a\\ {\log _6}24 = b\\ \end{array}\]

    aratati ca a(b + 3) – 3b + 1 = 0

    Se trec logaritmii in baza 2:
    a=\frac{\log_23+4}{2\log_23+3}, \ b=\frac{\log_23+3}{\log_23+1}
    Notam \log_23=x\Rightarrow a=\frac{x+4}{2x+3}, \ b=\frac{x+3}{x+1}

    Se scoate x in functie de a si se inlocuieste in expresia lui b si se obtine relatia ceruta.

  2. red12dog34
    veteran (III)

    marius9505 wrote: 1

    2) a = {\log _{\sin x}}\cos x\\ unde x apartine (0, pi/2)- {pi/4} exprimati in functie de a log in baza tg(x) din sin2x/2


    \log_{\tan x}\frac{\sin 2x}{2}=\frac{\log_{\sin x}(\sin x\cos x)}{\log_{\sin x}\frac{\sin x}{\cos x}}=\frac{1+\log_{\sin x}\cos x}{1-\log_{\sin x}\cos x}=\frac{1+a}{1-a}

  4. PhantomR
    expert (VI)

    La 4) 5. cred ca nu e bine ceva.

  5. marius9505

    da am vazut ca am gresit la 4
    la punctul a ) am scris gresit primul termen finn log in baza 2 din a
    la b) e lg nu log
    si c e corect fiin lg din radical de ordinul 2^n din a formula generala

    si la exercitiul 5 nu am gresit nimic asa e pb.

  6. marius9505

    Explica-mi si mie cum ai facut la 2

  8. andu_flavius95
    maestru (V)

    marius9505 wrote:

    10)
    aratati ca

        \[\frac{1}{{{{\log }_2}1 + {{\log }_2}2 + ... + {{\log }_2}n}} + \frac{1}{{{{\log }_3}1 + {{\log }_3}2 + ... + {{\log }_3}n}} + ... + \frac{1}{{{{\log }_n}1 + {{\log }_n}2 + ... + {{\log }_n}n}} = 1\]

        \[ \begin{array}{l} \frac{1}{{\log _2 1 + \log _2 2 + ... + \log _2 n}} + \frac{1}{{\log _3 1 + \log _3 2 + ... + \log _3 n}} + ... + \frac{1}{{\log _n 1 + \log _n 2 + ... + \log _n n}} = \\ = \frac{1}{{\log _2 (1 \cdot 2 \cdot ... \cdot n)}} + \frac{1}{{\log _3 (1 \cdot 2 \cdot ... \cdot n)}} + ... + \frac{1}{{\log _n (1 \cdot 2 \cdot ... \cdot n)}} = \\ = \log _{(1 \cdot 2 \cdot .. \cdot n)} 2 + \log _{(1 \cdot 2... \cdot n)} 3 + ... + \log _{(1 \cdot 2... \cdot n)} n = \\ = \log _{(1 \cdot 2... \cdot n)} (2 \cdot 3 \cdot ... \cdot n) = 1 \\ \end{array} \]

  9. andrei_g

    marius9505 wrote:
    8 )
    a)sa se arate ca daca

        \[\begin{array}{l} x = {\log _a}bc\\ y = {\log _b}ac\\ z = {\log _c}ab \end{array}\]

    x+y+z+2=xyz
    b) fie a,b,c apartine R*+ .Aratati ca daca

        \[\begin{array}{l} {2^a} = 35\\ {5^b} = 14\\ {7^c} = 10 \end{array}\]

    abc-(a+b+c)= 2

    Probleme acestea le-am intalnit si eu dar nu mi-au iesit…Cum se rezolva??

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