Funkcja logarytmiczna

[dotkatwoja]

Rozwiąż:
a) \(\frac{1}{5−4 \log x} + \frac{4}{1+\log x} = 3\),

b) \(\log_{16}x + \log_4 x + \log_2 x = 7\),

c) \(x^{\log x} = 100x\),

d) \(\log(2^x + 4^x) − \log 8 = \log(2^(x−1) −\frac{1}{4})\)

e)\(\log_{\frac{1}{4}} |x − 3| < −2,\)

f) \(\log_x\frac{1}{27} > −3\),

g)\(\log _x\frac{2x−1}{x−1} > 1\),

h)\(\log_{\frac{1}{3}}[\log_4(x^2 − 5)] > 0\),

i)\(|3 \log x − 1| < 2,\)

j) \(\log_{2x−3} (3x^2 − 7x + 3) < 2\)


Z góry dziękuję za pomoc.

[eresh]

Rozwiąż:
a) 1/(5−4 log x) + 4/(1+log x) = 3,

\(\frac{1}{5-4\log x}+\frac{4}{1+\log x}=3\\
x>0\\
5-4\log x\neq 0 \So -4\log x\neq -5\;\So \log x\neq \frac{5}{4}\So x\neq 10^{\frac{5}{4}}\\
1+\log x\neq 0\So \log x\neq -1\So x\neq 0,1\)

\(1+\log x+20-16\log x=3(5-4\log x)(1+\log x)\\
21-15\log x=15+15\log x-12\log x-12\log ^2x\\
12\log^2x-18\log x+6=0\\
\log x=t\\
12t^2-18t+6=0\\
t=1\So \log x=1\So x=10\\
t=\frac{1}{2}\So \log x=\frac{1}{2}\;\So x=\sqrt{10}\)

[eresh]

Rozwiąż:
c) x^log x = 100x,

\(x^{\log x}=100x\\
x>0\\
\log x^{\log x}=\log (100x)\\
\log x\log x=\log 100+\log x\\
\log^2x-\log x-2=0\\
\log x=t\\
t^2-t-2=0\\
t=2\;\So \log x=2\;\So x=100\\
t=-1\So \log x=-1\;\So x=0,1\)

[eresh]

Rozwiąż:
b) log\/16 x + log\/4 x + log\/2 x = 7,

\(\log_{16}x+\log_4x+\log_2x=7\\
x>0\\
\frac{\log_2x}{\log_216}+\frac{\log_2x}{\log_24}+\log_2x=7\\
\frac{\log_2x}{4}+\frac{\log_2x}{2}+\log_2x=7\\
\frac{7}{4}\log_2x=7\\
\log_2x=4\\
x=2^4\)

[eresh]

Rozwiąż:
d) log(2^x + 4^x) − log 8 = log(2^(x−1) − 1/4)

\(\log (2^x+4^x)-\log 8=\log (2^{x-1}-\frac{1}{4})\\\)
\(2^{x-1}-\frac{1}{4}>0\\
2^{x-1}>2^{-2}\\
x-1>-2\\
x>-1\)

\(\log\frac{2^x+2^{2x}}{8}=\log(2^{x-1}-\frac{1}{4})\\
\frac{2^x+2^{2x}}{8}=\frac{1}{2}\cdot 2^{x}-\frac{1}{4}\\
2^{x}+2^{2x}=4\cdot 2^x-2\\
2^{2x}-3\cdot 2^x+2=0\\
2^x=t, t>0\\
t^2-3t+2=0\\
t=1\So 2^x=1\So x=0\\
t=2\So 2^x=2\So x=1\)

[eresh]

Rozwiąż:

e)log \vee (1/4) |x − 3| < −2,

\(\log_{0,25}|x-3|<-2
x\neq 3\\
\log_{0,25}|x-3|<\log_{0,25}16\\
|x-3|>16\\
x-3>16\;\; \vee \;\;x-3<-16\\
x>19\;\;\vee\;\;x<-12\\
x\in (-\infty, -12)\cup (19,\infty)\)

[eresh]

Rozwiąż:


f) log \vee x 1/27 > −3,

\(\log_x\frac{1}{27}>-3\)

1. dla x>1
\(\log_x\frac{1}{27}>\log_xx^{-3}\\
\frac{1}{27}>\frac{1}{x^3}\\
\frac{1}{3}>\frac{1}{x}\\
x>3\)

2. dla \(x\in (0,1)\)
\(\log_x\frac{1}{27}>\log_xx^{-3}\\
\frac{1}{27}<\frac{1}{x^3}\\
\frac{1}{3}<\frac{1}{x}\\
x<3\\
x\in (0,1)\)

odp. \(x\in (0,1)\cup (3,\infty)\)

[eresh]

Rozwiąż:

g) log \vee x (2x−1)/(x−1) > 1,

\(\log_x\frac{2x-1}{x-1}>1\)

\(\frac{2x-1}{x-1}>0\\
(2x-1)(x-1)>0\\
x\in (-\infty, \frac{1}{2})\cup (1,\infty)\)

\(x>0\\
x\neq 1\\
D=(0,\frac{1}{2})\cup (1,\infty)\)

1. dla \(x\in (0,\frac{1}{2})\)
\(\log_x\frac{2x-1}{x-1}>1\\
\log_x\frac{2x-1}{x-1}>\log_xx\\
\frac{2x-1}{x-1}<x\\
\frac{2x-x-x^2+x}{x-1}<0\\
\frac{-x^2+2x}{x-1}<0\\
-x(x-2)(x-1)<0\\
x\in (0,\frac{1}{2})\)

2. dla \(x\in (1,\infty)\)
\(\log_x\frac{2x-1}{x-1}>1\\
\log_x\frac{2x-1}{x-1}>\log_xx\\
\frac{2x-1}{x-1}>x\\
\frac{2x-x-x^2+x}{x-1}>0\\
\frac{-x^2+2x}{x-1}>0\\
-x(x-2)(x-1)>0\\
x\in (1,2)\)


odp. \(x\in (0,\frac{1}{2})\cup (1,2)\)

[eresh]

Rozwiąż:

h) log \vee (1/3) [log4(x^2 − 5)] > 0,

\(\log_{\frac{1}{3}}[\log_4(x^2-5)]>0\)

\(\log_4(x^2-5)>0\;\;\; \wedge \;\;\;x^2-5>0\\
x^2-5>1\;\;\;\wedge\;\;\;(x-\sqrt{5})(x+\sqrt{5})>0\\
(x-\sqrt{6})(x+\sqrt{6})>0\;\;\wedge\;\;(x-\sqrt{5})(x+\sqrt{5})>0\\
D=(-\infty, -\sqrt{6})\cup (\sqrt{6},\infty)\)

\(\log_{\frac{1}{3}}[\log_4(x^2-5)]>0\\
\log_{\frac{1}{3}}[\log_4(x^2-5)]>\log_{\frac{1}{3}}1\\
\log_4(x^2-5)<1\\
\log_4(x^2-5)<\log_44\\
x^2-5<4\\
x^2-9<0\\
x\in (-\infty, -3)\cup (3\infty)\)

[eresh]

Rozwiąż:

i) |3 log x − 1| < 2,

\(|3\log x-1|<2\\
x>0\\\)
\(-2<3\log x-1<2\\
-1<3\log x<3\\
-\frac{1}{3}<\log x<1\\
\log 10^{-\frac{1}{3}}<\log x<\log 10\\
x\in (\frac{1}{\sqrt[3]{10}},10)\)

[eresh]

Rozwiąż:
j) log \vee (2x−3) (3x2 − 7x + 3) < 2

\(3x^2-7x+3>0\;\;\; \wedge \;\;\;2x-3>0\;\;\;\wedge\;\;\;2x-3\neq 1\\
x\in (-\infty, \frac{7-\sqrt{13}}{6})\cup (\frac{7+\sqrt{13}}{6},\infty)\;\;\wedge\;\;x>\frac{3}{2}\;\;\;\wedge\;\;\;x\neq 2\\
D=(\frac{3}{2},2)\cup (2,\infty)\)

1. dla \(x\in (\frac{3}{2},2)\)
\(\log_{2x-3}(3x^2-7x+3)<2\\
\log_{2x-3}(3x^2-7x+3)<\log_{2x-3}(2x-3)^2\\
3x^2-7x+3>4x^2-12x+9\\
-x^2+5x-6>0\\
\emptyset\)

2. dla \(x\in (2,\infty)\)
\(\log_{2x-3}(3x^2-7x+3)<2\\
\log_{2x-3}(3x^2-7x+3)<\log_{2x-3}(2x-3)^2\\
3x^2-7x+3<4x^2-12x+9\\
-x^2+5x-6<0\\
x\in (3,\infty)\)