$\log_{3}(2x+3)-\log_{3}(x-2)=2$
$\log_{3}\frac{2x+3}{x-2}=\log_{3}9$
$\frac{2x+3}{x-2}=9$
$2x+3=9x-18$
$7x=21$
$x=3$
$2x+3>0$
$x>-\frac{3}{2}$
$x-2>0$
$x>2$
$K=\left \{ 3 \right \}$
Zdroj: PETÁKOVÁ, Jindra. Matematika - příprava k maturitě a k přijímacím zkouškám na vysoké školy. Praha: Prometheus, 1998. ISBN 80-7196-099-3.
$\large \log_{2}(x+1)=3$
$\log_{2}(x+1)=3\log_{2}2$
$\log_{2}(x+1)=\log_{2}2^{3}$
$x+1=8$
$x=7$
$x>-1$
$K=\left \{ 7 \right \}$
$\large 4\log_{3}(2x-1)=12$
$4\log_{3}(2x-1)=12\log_{3}3$
$\log_{3}(2x-1)=3\log_{3}3$
$\log_{3}(2x-1)=\log_{3}3^{3}$
$2x-1=27$
$x=14$
$x>-\frac{1}{2}$
$K=\left \{ 14 \right \}$
$\large \log_{\frac{1}{2}}(2-x)=-2$
$\log_{\frac{1}{2}}(2-x)=-2\log_{\frac{1}{2}}\frac{1}{2}$
$\log_{\frac{1}{2}}(2-x)=\log_{\frac{1}{2}}\frac{1}{2}^{-2}$
$\log_{\frac{1}{2}}(2-x)=\log_{\frac{1}{2}}4$
$2-x=4$
$x=-2$
$x<2$
$K=\left \{ -2 \right \}$
$\large \log_{4}(5x-4)=2$
$\log_{4}(5x-4)=2\log_{4}4$
$\log_{4}(5x-4)=\log_{4}4^{2}$
$\log_{4}(5x-4)=\log_{4}16$
$5x-4=16$
$x=4$
$x>\frac{4}{5}$
$K=\left \{ 4 \right \}$
$\large \log_{2}\log_{3}\log_{\frac{1}{2}}x=0$
$ \log_{2}a=0$
$a=1$
$\log_{3}\log_{\frac{1}{2}}x=1$
$\log_{3}b=1$
$b=3$
$\log_{\frac{1}{2}}x=3$
$x=\frac{1}{8}$
$x>0$
$K=\left \{ \frac{1}{8} \right \}$
$\large \log_{\frac{1}{2}}\log_{3}(1+20\log_{2}x)=-2$
$ \log_{\frac{1}{2}}a=-2$
$a=4$
$\log_{3}(1+20\log_{2}x)=4$
$\log_{3}b=4$
$b=81$
$1+20\log_{2}x=81$
$20\log_{2}x=80$
$\log_{2}x=4$
$x=16$
$x>0$
$K=\left \{ 16 \right \}$
$\large \log_{2}(14+2\log_{7}(1+2\log_{\frac{1}{2}}x))=4$
$14+2\log_{7}(1+2\log_{\frac{1}{2}}x)=16$
$\log_{7}(1+2\log_{\frac{1}{2}}x)=1$
$1+2\log_{\frac{1}{2}}x=7$
$\log_{\frac{1}{2}}x=3$
$x=\frac{1}{8}$
$x>0$
$K=\left \{ \frac{1}{8} \right \}$
$\large \log_{9}(3\log_{2}(1+\log_{3}(1-2\log_{3}x)))=\frac{1}{2}$
$3\log_{2}(1+\log_{3}(1-2\log_{3}x))=3$
$3\log_{2}(1+\log_{3}(1-2\log_{3}x))=3$
$1+\log_{3}(1-2\log_{3}x))=2$
$\log_{3}(1-2\log_{3}x))=1$
$1-2\log_{3}x=3$
$-2\log_{3}x=2$
$-1\log_{3}x=1$
$x=\frac{1}{3}$
$x>0$
$K=\left \{ \frac{1}{3} \right \}$
$\large \log_{5}(x^{2}+2x)=\log_{5}(-3x)$
$x^{2}+2x=-3x$
$x^{2}+5x=0$
$x(x+5)=0$
$x_{1}=0\;x_{2}=-5$
$x<-2$
$K=\left \{ -5 \right \}$
$\large \log x^{2}=\log(4-x)^{2}$
$x^{2}=(4-x)^{2}$
$2x^{2}-4=0$
$2(x^{2}-2)=0$
$x_{1}=\sqrt{2}\;x_{2}=-\sqrt{2}$
$x>-2\;\wedge\;x<2$
$K=\left \{ \pm \sqrt{2} \right \}$
$\large \log_{0,1}(x^{2}-5x)=\log_{0,1}(5x+11)$
$x^{2}-5x=5x+11$
$x^{2}-10x-11=0$
$(x-11)(x+1)=0$
$x_{1}=11\;x_{2}=-1$
$K=\left \{ 11;-1 \right \}$
$\large \log_{2}(x^{2}-x)=\log_{2}x$
$x^{2}-x=x$
$x^{2}-2x=0$
$x(x-2)=0$
$x_{1}=0\;x_{2}=2$
$K=\left \{ 2 \right \}$
$\large \log x=2\log5+\log4$
$\log x=\log5x^{2} \times 4$
$\log x=\log 100$
$\log x=100$
$K=\left \{ 100 \right \}$
$\large \frac{\log_{3}x}{1+\log_{3}2} =2$
$\log_{3}x =2(1+\log_{3}2)$
$\log_{3}x =2+2\log_{3}2)$
$\log_{3}x =\log_{3}9+\log_{3}4$
$\log_{3}x =\log_{3}36$
$x=36$
$K=\left \{ 36 \right \}$
$\large \log_{6}(x+1)+\log_{6}x=1$
$\log_{6}(x+1)x=\log_{6}6$
$x(x+1)=6$
$x^{2}+x-6=0$
$(x+3)(x-2)=0$
$x_{1}=-3\;x_{2}=2$
$K=\left \{ 2 \right \}$
$\large \log_{2}(x+7)-\log_{2}x=3$
$\log_{2} \frac{x+7}{x}=\log_{2}8$
$\frac{x+7}{x}=8$
$x+7=8x$
$-7x+7=0$
$-7(x-1)=0$
$x=1$
$K=\left \{1 \right \}$
$\large \log (x+3)=\log x+\log3$
$x+3=3x$
$2x=3$
$x=\frac{3}{2}$
$K=\left \{\frac{3}{2} \right \}$
$\large \log_{8} \sqrt{x+30}+\log_{8}\sqrt{x}=1$
$\log_{8} \sqrt{x+30} \times \sqrt{x}=\log_{8}8$
$\sqrt{x(x+30)} = 8$
$x(x+30) = 64$
$x^{2}+30x-64=0$
$(x-2)(x+32)=0$
$x_{1}=2\;x_{2}=-32$
$K=\left \{ 2 \right \}$
$\large \log x^{5}-\log x^{4}+\log x^{3}=12$
$ 5\log x-4\log x+3\log x=12$
$ 4\log x=12$
$ \log x=3$
$x=1000$
$K=\left \{1000 \right \}$
$\large \log\sqrt{x}+\log\frac{1}{x^{2}}-\log x^{3} + \frac{11}{2}=\frac{\log x ^{2}}{1+\log 10}$
$\log x^{\frac{1}{2}}+\log x^{-2}-\log x^{3}+ \frac{11}{2}=\frac{\log x ^{2}}{2}$
$\frac{1}{2}\log x-2\log x-3\log x+ \frac{11}{2}=\frac{2\log x}{2}$
$\log x-4\log x-6\log x+ 11=2\log x$
$11\log x=11$
$\log x=1$
$x=10$
$K=\left \{10 \right \}$
$\large 3\log 2x^{2}+2\log3 x^{3}=5\log x+2\log6 x^{2}$
$\log (2x^{2})^{3}+\log (3 x^{3})^{2}=\log x^{5}+\log (6 x^{2})^{2}$
$\log 8x^{6} \times 9x^{6}=\log x^{5} \times 36x^{6}$
$72x^{12}=36x^{11}$
$2x=1$
$x=\frac{1}{2}$
$K=\left \{\frac{1}{2} \right \}$
$\large 0,5(3\log5-1-\log x)=2-\log5$
$3\log5-1-\log x=4-2\log5$
$5\log5-\log x=5$
$\log x^{-1}=\log10^{5}-log5^{5}$
$x^{-1}=(\frac{10}{5})^{5}$
$x=\frac{1}{32}$
$K=\left \{\frac{1}{32} \right \}$
$\large \log_{4}(3x+2)-2\log_{4}x=2-\log_{4}8$
$\log_{4}\frac{3x+2}{x^{2}}=\log_{4}16-\log_{4}8$
$\log_{4}\frac{3x+2}{x^{2}}=\log_{4}\frac{16}{8}$
$\frac{3x+2}{x^{2}}=\frac{16}{8}$
$\frac{3x+2}{x^{2}}=2$
$3x+2=2x^{2}$
$2x^{2}-3x-2=0$
$x^{2}-1,5x-1=0$
$(x-2)(x+\frac{1}{2})=0$
$x_{1}=2\;x_{2}=-\frac{1}{2}$
$K=\left \{ 2 \right \}$
$\large 1+ \log_{3}(5-x)-\log_{3}(2x-1)=\log_{3}(2x-1)$
$\log_{3} 3 + \log_{3}(5-x)-\log_{3}(2x-1)=\log_{3}(2x-1)$
$\log_{3} \frac{3(5-x)}{2x-1}=\log_{3}(2x-1)$
$\frac{3(5-x)}{2x-1}=(2x-1)$
$3(5-x)=(2x-1)x^{2}$
$15-3x=4x^{2}-4x+1$
$4x^{2}-x-14=0$
$x^{2}-\frac{1}{4}x-\frac{14}{4}=0$
$(x-2)(x+\frac{7}{4})=0$
$x_{1}=2\;x_{2}=-\frac{7}{4}$
$K=\left \{ 2 \right \}$
$\large \log_{2} \frac{3-x}{x+3}=-2$
$\log_{2} \frac{3-x}{x+3}=\log_{2} \frac{1}{4}$
$\frac{3-x}{x+3}=\frac{1}{4}$
$4(3-x)=x+3$
$12-4x=x+3$
$-5x=-9$
$x=\frac{9}{5}$
$K=\left \{1\frac{4}{5} \right \}$
$\large \log_{3} \frac{6x-2}{x-3}=2$
$\log_{3} \frac{6x-2}{x-3}=\log_{3} 9$
$\frac{6x-2}{x-3}=9$
$6x-2=9x-27$
$-3x=-25$
$x=\frac{25}{3}$
$K=\left \{8\frac{1}{3} \right \}$
$\large \log_{\frac{1}{2}} \frac{x}{x+14}=\frac{\log 125}{\log 5}$
$\log_{\frac{1}{2}} \frac{x}{x+14}=3$
$\log_{\frac{1}{2}} \frac{x}{x+14}=\log_{\frac{1}{2}} \frac{1}{8}$
$\frac{x}{x+14}=\frac{1}{8}$
$8x=x+14$
$x=2$
$K=\left \{2 \right \}$
$\large \log_{7} \frac{x^{2}+1}{x^{2}-1}=1$
$\log_{7} \frac{x^{2}+1}{x^{2}-1}=\log_{7} 7$
$\frac{x^{2}+1}{x^{2}-1}=7$
$x^{2}+1=7x^{2}-7$
$6x^{2}-8=0$
$2(3x^{2}-4)=0$
$x^{2}=\frac{4}{3}$
$x_{1}=2\frac{\sqrt{3}}{3}\;x_{2}=-2\frac{\sqrt{3}}{3}$
$K=\left \{2\frac{\sqrt{3}}{3},\;-2\frac{\sqrt{3}}{3} \right \}$
$\large \frac {\log_{3} (6x-2)}{\log_{3} (x-3)}=2$
$\log_{3} (6x-2)=\log_{3} (x-3)^{2}$
$(6x-2)=(x-3)^{2}$
$x^{2}-12x+11=0$
$(x-11)(x-1)=0$
$x_{1}=11\;x_{2}=1$
$K=\left \{ 11 \right \}$
$\large \frac {\log_{5} (x-\frac{1}{4})}{\log_{5} (x+\frac{7}{2})}=-1$
$\log_{5} (x-\frac{1}{4})=-1\log_{5} (x+\frac{7}{2})$
$(x-\frac{1}{4})=\frac{1}{x+\frac{7}{2}}$
$(x-\frac{1}{4})(x+\frac{7}{2})=1$
$x^{2}+\frac{7x}{2}-\frac{x}{4}-\frac{7}{8}-1=0$
$8x^{2}+28x-2x-7-8=0$
$8x^{2}+26x-15=0$
$x_{1}=\frac{1}{2}\;x_{2}=-\frac{15}{4}$
$K=\left \{ \frac{1}{2} \right \}$
$\large \frac {2\log 3x}{\log (2-7x)}=1$
$2\log 3x=\log (2-7x)$
$9x^{2}+7x-2=0$
$x_{1}=\frac{2}{9}\;x_{2}=-1$
$K=\left \{ \frac{2}{9} \right \}$
$\large \frac {\log x}{\log (x-2)}=\frac {\log 9}{\log 3}$
$\frac {\log x}{\log (x-2)}=2$
$\log x=2\log (x-2)$
$x=(x-2)^{2}$
$x^{2}-5x+4=0$
$(x-4)(x-1)=0$
$x_{1}=4\;x_{2}=1$
$K=\left \{ 4 \right \}$
$\large \log_{2}^{2}x+2\log_{2}x-3=0$
$S:\log_{2}x=y$
$y^{2}+2y-3=0$
$(y+3)(y-1)=0$
$y_{1}=-3\;y_{2}=1$
$x_{1}:\log_{2}x=-3$
$x_{1}=\frac{1}{8}$
$x_{2}:\log_{2}x=1$
$x_{2}=2$
$K=\left \{\frac{1}{8};2 \right \}$
$\large 4 \log_{9} x (\log_{9}x-1)=2+3\log_{9} x$
$S:\log_{9}x=y$
$4y^{2}-7y-2=0$
$(y-2)(y+\frac{1}{4})=0$
$y_{1}=2\;y_{2}=-\frac{1}{4}$
$x_{1}:\log_{9}x=2$
$x_{1}=81$
$x_{2}:\log_{9}x=-\frac{1}{4}$
$x_{2}=\frac{\sqrt{3}}{3}$
$K=\left \{\frac{\sqrt{3}}{3};81 \right \}$
$\large \log_{\frac{1}{2}} (x+1) + 5 \log_{\frac{1}{2}} (x+1) = 6$
$S:\log_{\frac{1}{2}}(x+1)=y$
$y^{2}+5y-6=0$
$(y+6)(y-1)=0$
$y_{1}=-6\;y_{2}=1$
$x_{1}:\log_{\frac{1}{2}}(x+1)=-6$
$x_{1}=63$
$x_{2}:\log_{\frac{1}{2}}(x+1)=1$
$x_{2}=-\frac{1}{2}$
$K=\left \{-\frac{1}{2};63 \right \}$
$\large 4 \log_{3} (2x+1) + \log_{3} \sqrt{2x+1} = \frac{3}{2}\log_{3}^{2} (2x+1) - 6$
$4 \log_{3} (2x+1) + \frac{1}{2} \log_{3} {2x+1} = \frac{3}{2}\log_{3}^{2} (2x+1) - 6$
$\frac{3}{2}\log_{3}^{2} (2x+1) -4,5 \log_{3} (2x+1) - 6 = 0$
$S:\log_{3}(2x+1)=y$
$\frac{3}{2}y^{2}-4,5y-6=0$
$3y^{2}-9y-12=0$
$y^{2}-3y-4=0$
$(y-4)(y+1)=0$
$y_{1}=4\;y_{2}=-1$
$x_{1}:\log_{3}(2x+1)=4$
$x_{1}:\log_{3}(2x+1)=\log_{3}81$
$x_{1}:2x=80$
$x_{1}=40$
$x_{2}:\log_{3}(2x+1)=-1$
$x_{2}:\log_{3}(2x+1)=\log_{3} \frac{1}{3}$
$x_{2}:2x=- \frac{2}{3}$
$x_{2}=-\frac{1}{3}$
$K=\left \{-\frac{1}{3};40 \right \}$
Další příklady:
Stanovte podmínky a rešte v $R$ logaritmické rovnice:
a) $\log x-\log (2 x-5)=\log 3$
b) $\log (2 x-3)-\log (x+1)=-\log 3$
c) $\log (3 x-1)-\log 5=1$
d) $\log x=-\log 2$
e) $2 \log x=3 \log 4$
f) $\log 8 x+\log 3 x=\log 48$
g) $\log 16 x-\log 2 x+\log 3 x=\log 9+\log 4-\log 6$
h) $\log (x+2)+\log (x-7)=2 \log (x-4)$
i) $\log 5 x+\log (2 x+3)=1+2 \log (3-x)$
j) $\log (x+24)+\log (x-24)=2$
k) $\log (x+6)-\log x=\log 2-\log (x+1)$
l) $\log (x-3)+\log (x+3)=2 \log (3-x)$
m) $\log (x+1)+\log (x-1)-\log x=\log (x+2)$
n) $1+\log _{3}(5-x)-\log _{3}(2 x-1)=\log _{3}(2 x-1)$
o) $\frac{3+\log x}{2-\log x}=4$
p) $\frac{\log \left(x^{2}+3\right)}{\log (x+3)}=2$
q) $\frac{3}{2+\log x}-\frac{2}{1+\log x}=\frac{1}{10+\log x}$
r) $\frac{1+\log x}{\log x}-1=\frac{1-\log x}{\log x}$
s) $\frac{10+\log x}{7+\log x}=\frac{1}{7+\log x}+2$
t) $\log x+\frac{1}{\log x}=2$
u) $\log \sqrt{x}+\log \frac{1}{x^{2}}-\log x^{3}+\frac{11}{2}=\frac{\log x^{2}}{1+\log 10}$
v) $0,5(3 \log 5-1-\log x)=2-\log 5$
w) $\log _{4}(3 x+2)-2 \log _{4} x=2-\log _{4} 8$
x) $\log _{0,5}^{2}(x+1)+5 \log _{0,5}(x+1)=6$
y) $\log _{8} \sqrt{x+30}+\log _{8} \sqrt{x}=1$
z) $4 \log _{9} x\left(\log _{9} x-1\right)=2+3 \log _{9} x$
Výsledky:
a) $x \in\left(\frac{5}{2} ; \infty\right), K=\{3\}$;
b) $x \in\left(\frac{3}{2} ; \infty\right), K=\{2\} ;$
c) $x \in\left(\frac{1}{3} ; \infty\right), K=\{17\} ;$
d) $x \in(0 ; \infty) ; K=\left\{\frac{1}{2}\right\}$;
e) $x \in(0 ; \infty), K=\{8\}$;
f) $x \in(0 ; \infty), K=\{\sqrt{2}\}$;
g) $x \in(0 ; \infty), K=\left\{\frac{1}{4}\right\}$;
h) $x \in(7 ; \infty), K=\{10\}$;
i) $x \in(0 ; 3), K=\left\{\frac{6}{5}\right\}$;
j) $x \in(24 ; \infty), K=\{26\} ;$
k) $x \in(0 ; \infty), K=\varnothing$ ;
l) $x \in \varnothing, K=\varnothing$;
m) $x \in(1 ; \infty), K=\varnothing ;$
n) $x \in\left(\frac{1}{2} ; 5\right) ; K=\{2\}$;
o) $x \in(0 ; \infty)-\{100\} ; K=\{10\}$;
p) $x \in(-3 ; \infty) ; K=\{-1\}$;
q) $x \in(0 ; \infty)-\left\{10^{-10} ; 10^{-2} ; 10^{-1}\right\} ; K=\{100\} ;$
r) $x \in(0 ; \infty)-\{1\} ; K=\varnothing$ ;
s) $x \in(0 ; \infty)-\left\{10^{-7} ; 1\right\}, K=\left\{10^{-5}\right\}$;
t) $x \in(0 ; \infty)-\{1\} ; K=\{10\} ;$
u) $x \in(0 ; \infty) ; K=\{10\}$ ;
v) $x \in(0 ; \infty) ; K=\left\{\frac{1}{32}\right\} $;
w) $x \in(0 ; \infty) ; K=\{2\}$;
х) $x \in(0 ; \infty) ; K=\left\{-\frac{1}{2} ; 63\right\}$;
y) $x \in(0 ; \infty) ; K=\{2\}$ ;
z) $x \in(0 ; \infty) ; K=\left\{\frac{\sqrt{3}}{3} ; 81\right\}$