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User:Lmov
log
2
x
2
x
log
2
2
x
+
log
2
4
x
=
1
{\displaystyle \log _{2x}{\frac {2}{x}}\log _{2}^{2}x+\log _{2}^{4}x=1}
x
∈
(
0
;
1
2
)
∪
(
1
2
;
+
∞
)
{\displaystyle x\in \left(0;{\frac {1}{2}}\right)\cup \left({\frac {1}{2}};+\infty \right)}
ln
2
x
ln
2
x
ln
2
x
ln
2
2
+
ln
4
x
ln
4
2
=
1
{\displaystyle {\frac {\ln {\frac {2}{x}}\ln ^{2}x}{\ln {2x}\ln ^{2}2}}+{\frac {\ln ^{4}x}{\ln ^{4}2}}=1}
ln
2
x
ln
2
x
ln
2
2
+
ln
4
x
ln
2
x
=
ln
2
x
ln
4
2
{\displaystyle \ln {\frac {2}{x}}\ln ^{2}x\ln ^{2}2+\ln ^{4}x\ln {2x}=\ln {2x}\ln ^{4}2}
(
ln
2
−
ln
x
)
ln
2
x
ln
2
2
+
ln
4
x
(
ln
2
+
ln
x
)
−
(
ln
2
+
ln
x
)
ln
4
2
=
0
{\displaystyle \left(\ln 2-\ln x\right)\ln ^{2}x\ln ^{2}2+\ln ^{4}x\left(\ln 2+\ln x\right)-\left(\ln 2+\ln x\right)\ln ^{4}2=0}
ln
5
x
+
ln
4
x
ln
2
−
ln
3
x
ln
2
2
+
ln
2
x
ln
3
2
−
ln
x
ln
4
2
−
ln
5
2
=
0
{\displaystyle \ln ^{5}x+\ln ^{4}x\ln 2-\ln ^{3}x\ln ^{2}2+\ln ^{2}x\ln ^{3}2-\ln x\ln ^{4}2-\ln ^{5}2=0}
(
ln
x
ln
2
)
5
+
(
ln
x
ln
2
)
4
−
(
ln
x
ln
2
)
3
+
(
ln
x
ln
2
)
2
−
(
ln
x
ln
2
)
−
1
=
0
{\displaystyle \left({\frac {\ln x}{\ln 2}}\right)^{5}+\left({\frac {\ln x}{\ln 2}}\right)^{4}-\left({\frac {\ln x}{\ln 2}}\right)^{3}+\left({\frac {\ln x}{\ln 2}}\right)^{2}-\left({\frac {\ln x}{\ln 2}}\right)-1=0}
log
2
5
x
+
log
2
4
x
−
log
2
3
x
+
log
2
2
x
−
log
2
x
−
1
=
0
{\displaystyle \log _{2}^{5}x+\log _{2}^{4}x-\log _{2}^{3}x+\log _{2}^{2}x-\log _{2}x-1=0}
Using
Ruffini's rule
find that 1 is a root:
1
1
−
1
1
−
1
−
1
1
[
1
2
1
2
1
]
0
{\displaystyle {\begin{matrix}&1&1&-1&1&-1&-1\\1&\left[1\right.&2&1&2&\left.1\right]&0\end{matrix}}}
(
log
2
x
−
1
)
(
log
2
4
x
+
2
log
2
3
x
+
log
2
2
x
+
2
log
2
x
+
1
)
=
0
{\displaystyle \left(\log _{2}x-1\right)\left(\log _{2}^{4}x+2\log _{2}^{3}x+\log _{2}^{2}x+2\log _{2}x+1\right)=0}
log
2
x
=
1
{\displaystyle \log _{2}x=1}
x
1
=
2
(1)
{\displaystyle x_{1}=2{\mbox{ (1)}}}
log
2
4
x
+
2
log
2
3
x
+
log
2
2
x
+
2
log
2
x
+
1
=
0
{\displaystyle \log _{2}^{4}x+2\log _{2}^{3}x+\log _{2}^{2}x+2\log _{2}x+1=0}
log
2
2
x
+
2
log
2
x
+
1
+
2
1
log
2
x
+
1
log
2
2
x
=
0
{\displaystyle \log _{2}^{2}x+2\log _{2}x+1+2{\frac {1}{\log _{2}x}}+{\frac {1}{\log _{2}^{2}x}}=0}
(
log
2
2
x
+
1
log
2
2
x
)
+
2
(
log
2
x
+
1
log
2
x
)
+
1
=
0
{\displaystyle \left(\log _{2}^{2}x+{\frac {1}{\log _{2}^{2}x}}\right)+2\left(\log _{2}x+{\frac {1}{\log _{2}x}}\right)+1=0}
Let
y
=
log
2
x
+
1
log
2
x
⇒
y
2
−
2
=
log
2
2
x
+
1
log
2
2
x
{\displaystyle {\mbox{Let }}y=\log _{2}x+{\frac {1}{\log _{2}x}}\Rightarrow y^{2}-2=\log _{2}^{2}x+{\frac {1}{\log _{2}^{2}x}}}
y
2
−
2
+
2
y
+
1
=
0
{\displaystyle y^{2}-2+2y+1=0}
y
2
+
2
y
−
1
=
0
{\displaystyle y^{2}+2y-1=0}
y
1
,
2
=
−
1
±
2
{\displaystyle y_{1,2}=-1\pm {\sqrt {2}}}
log
2
x
+
1
log
2
x
=
−
1
−
2
{\displaystyle \log _{2}x+{\frac {1}{\log _{2}x}}=-1-{\sqrt {2}}}
log
2
2
x
+
(
2
+
1
)
log
2
x
+
1
=
0
{\displaystyle \log _{2}^{2}x+\left({\sqrt {2}}+1\right)\log _{2}x+1=0}
log
2
x
=
−
(
2
+
1
)
±
2
2
−
1
2
{\displaystyle \log _{2}x={\frac {-\left({\sqrt {2}}+1\right)\pm {\sqrt {2{\sqrt {2}}-1}}}{2}}}
x
2
,
3
=
2
−
(
2
+
1
)
±
2
2
−
1
2
(2)
{\displaystyle x_{2,3}=2^{\frac {-\left({\sqrt {2}}+1\right)\pm {\sqrt {2{\sqrt {2}}-1}}}{2}}{\mbox{ (2)}}}
log
2
x
+
1
log
2
x
=
−
1
+
2
{\displaystyle \log _{2}x+{\frac {1}{\log _{2}x}}=-1+{\sqrt {2}}}
log
2
2
x
+
(
1
−
2
)
log
2
x
+
1
=
0
{\displaystyle \log _{2}^{2}x+\left(1-{\sqrt {2}}\right)\log _{2}x+1=0}
D
=
(
1
−
2
)
2
−
4
=
−
1
−
2
2
<
0
⇒
roots are complex
{\displaystyle D=\left(1-{\sqrt {2}}\right)^{2}-4=-1-2{\sqrt {2}}<0\Rightarrow {\mbox{roots are complex}}}
Thus, from (1) and (2):
x
1
=
2
,
x
2
,
3
=
2
−
(
2
+
1
)
±
2
2
−
1
2
{\displaystyle x_{1}=2,x_{2,3}=2^{\frac {-\left({\sqrt {2}}+1\right)\pm {\sqrt {2{\sqrt {2}}-1}}}{2}}}
![{\displaystyle {\begin{matrix}&1&1&-1&1&-1&-1\\1&\left[1\right.&2&1&2&\left.1\right]&0\end{matrix}}}](https://cdn.statically.io/img/wikimedia.org/api/rest_v1/media/math/render/svg/035f241d4fd425631dc2e117c08b5de5ba5f74e4)
