116957
If \(x>0\) and \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\), then value of \(x\) is
1 4
2 64
3 8
4 2
Explanation:
A Given \(x>0\) then \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\). this can be written as \(\log _4 x^2(x+1)-\log _4(x+1)=2\) or \(\quad \log _4 x^2+\log _4(x+1)-\log _4(x+1)=2\) \(\log _4 x^2=2\) \(\mathrm{x}^2=4^2\) \(\therefore \mathrm{x}=4\)
EAMCET-1993
Sets, Relation and Function
116958
The real value(s) of \(x\) which satisfy \((5+2 \sqrt{6})^{x^2-3}+(5-2 \sqrt{6})^{x^2-3}=10\) is/are
1 \(2,-\sqrt{2}\)
2 \(\pm 2, \pm \sqrt{2}\)
3 \(2, \sqrt{2}\)
4 \(-2,-\sqrt{2}\)
Explanation:
B Given, \((5+2 \sqrt{6})^{\mathrm{x}^2-3}+(5-2 \sqrt{3})^{\mathrm{x}^2-3}=10\) Let \(t=(5+2 \sqrt{6})^{x^2-3}\) then \(\frac{1}{t}=(5-2 \sqrt{6})^{x^2-3}\) Then we have \(t+\frac{1}{t}=10\) Again \((5+2 \sqrt{6})^{x^2-3}=(5-2 \sqrt{6})\) \(=\frac{1}{5+2 \sqrt{6}}\) \(=(5+2 \sqrt{6})^{-1}\) \(\therefore x^2-3=-1\) \(\text { or } x^2=2\) \(\text { or } x= \pm \sqrt{2}\)
EAMCET-1992
Sets, Relation and Function
116960
The value of \(\left(\frac{1}{\log _3 12}+\frac{1}{\log _4 12}\right)\) is
116957
If \(x>0\) and \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\), then value of \(x\) is
1 4
2 64
3 8
4 2
Explanation:
A Given \(x>0\) then \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\). this can be written as \(\log _4 x^2(x+1)-\log _4(x+1)=2\) or \(\quad \log _4 x^2+\log _4(x+1)-\log _4(x+1)=2\) \(\log _4 x^2=2\) \(\mathrm{x}^2=4^2\) \(\therefore \mathrm{x}=4\)
EAMCET-1993
Sets, Relation and Function
116958
The real value(s) of \(x\) which satisfy \((5+2 \sqrt{6})^{x^2-3}+(5-2 \sqrt{6})^{x^2-3}=10\) is/are
1 \(2,-\sqrt{2}\)
2 \(\pm 2, \pm \sqrt{2}\)
3 \(2, \sqrt{2}\)
4 \(-2,-\sqrt{2}\)
Explanation:
B Given, \((5+2 \sqrt{6})^{\mathrm{x}^2-3}+(5-2 \sqrt{3})^{\mathrm{x}^2-3}=10\) Let \(t=(5+2 \sqrt{6})^{x^2-3}\) then \(\frac{1}{t}=(5-2 \sqrt{6})^{x^2-3}\) Then we have \(t+\frac{1}{t}=10\) Again \((5+2 \sqrt{6})^{x^2-3}=(5-2 \sqrt{6})\) \(=\frac{1}{5+2 \sqrt{6}}\) \(=(5+2 \sqrt{6})^{-1}\) \(\therefore x^2-3=-1\) \(\text { or } x^2=2\) \(\text { or } x= \pm \sqrt{2}\)
EAMCET-1992
Sets, Relation and Function
116960
The value of \(\left(\frac{1}{\log _3 12}+\frac{1}{\log _4 12}\right)\) is
116957
If \(x>0\) and \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\), then value of \(x\) is
1 4
2 64
3 8
4 2
Explanation:
A Given \(x>0\) then \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\). this can be written as \(\log _4 x^2(x+1)-\log _4(x+1)=2\) or \(\quad \log _4 x^2+\log _4(x+1)-\log _4(x+1)=2\) \(\log _4 x^2=2\) \(\mathrm{x}^2=4^2\) \(\therefore \mathrm{x}=4\)
EAMCET-1993
Sets, Relation and Function
116958
The real value(s) of \(x\) which satisfy \((5+2 \sqrt{6})^{x^2-3}+(5-2 \sqrt{6})^{x^2-3}=10\) is/are
1 \(2,-\sqrt{2}\)
2 \(\pm 2, \pm \sqrt{2}\)
3 \(2, \sqrt{2}\)
4 \(-2,-\sqrt{2}\)
Explanation:
B Given, \((5+2 \sqrt{6})^{\mathrm{x}^2-3}+(5-2 \sqrt{3})^{\mathrm{x}^2-3}=10\) Let \(t=(5+2 \sqrt{6})^{x^2-3}\) then \(\frac{1}{t}=(5-2 \sqrt{6})^{x^2-3}\) Then we have \(t+\frac{1}{t}=10\) Again \((5+2 \sqrt{6})^{x^2-3}=(5-2 \sqrt{6})\) \(=\frac{1}{5+2 \sqrt{6}}\) \(=(5+2 \sqrt{6})^{-1}\) \(\therefore x^2-3=-1\) \(\text { or } x^2=2\) \(\text { or } x= \pm \sqrt{2}\)
EAMCET-1992
Sets, Relation and Function
116960
The value of \(\left(\frac{1}{\log _3 12}+\frac{1}{\log _4 12}\right)\) is
116957
If \(x>0\) and \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\), then value of \(x\) is
1 4
2 64
3 8
4 2
Explanation:
A Given \(x>0\) then \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\). this can be written as \(\log _4 x^2(x+1)-\log _4(x+1)=2\) or \(\quad \log _4 x^2+\log _4(x+1)-\log _4(x+1)=2\) \(\log _4 x^2=2\) \(\mathrm{x}^2=4^2\) \(\therefore \mathrm{x}=4\)
EAMCET-1993
Sets, Relation and Function
116958
The real value(s) of \(x\) which satisfy \((5+2 \sqrt{6})^{x^2-3}+(5-2 \sqrt{6})^{x^2-3}=10\) is/are
1 \(2,-\sqrt{2}\)
2 \(\pm 2, \pm \sqrt{2}\)
3 \(2, \sqrt{2}\)
4 \(-2,-\sqrt{2}\)
Explanation:
B Given, \((5+2 \sqrt{6})^{\mathrm{x}^2-3}+(5-2 \sqrt{3})^{\mathrm{x}^2-3}=10\) Let \(t=(5+2 \sqrt{6})^{x^2-3}\) then \(\frac{1}{t}=(5-2 \sqrt{6})^{x^2-3}\) Then we have \(t+\frac{1}{t}=10\) Again \((5+2 \sqrt{6})^{x^2-3}=(5-2 \sqrt{6})\) \(=\frac{1}{5+2 \sqrt{6}}\) \(=(5+2 \sqrt{6})^{-1}\) \(\therefore x^2-3=-1\) \(\text { or } x^2=2\) \(\text { or } x= \pm \sqrt{2}\)
EAMCET-1992
Sets, Relation and Function
116960
The value of \(\left(\frac{1}{\log _3 12}+\frac{1}{\log _4 12}\right)\) is
116957
If \(x>0\) and \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\), then value of \(x\) is
1 4
2 64
3 8
4 2
Explanation:
A Given \(x>0\) then \(\log _4\left(x^3+x^2\right)-\log _4(x+1)=2\). this can be written as \(\log _4 x^2(x+1)-\log _4(x+1)=2\) or \(\quad \log _4 x^2+\log _4(x+1)-\log _4(x+1)=2\) \(\log _4 x^2=2\) \(\mathrm{x}^2=4^2\) \(\therefore \mathrm{x}=4\)
EAMCET-1993
Sets, Relation and Function
116958
The real value(s) of \(x\) which satisfy \((5+2 \sqrt{6})^{x^2-3}+(5-2 \sqrt{6})^{x^2-3}=10\) is/are
1 \(2,-\sqrt{2}\)
2 \(\pm 2, \pm \sqrt{2}\)
3 \(2, \sqrt{2}\)
4 \(-2,-\sqrt{2}\)
Explanation:
B Given, \((5+2 \sqrt{6})^{\mathrm{x}^2-3}+(5-2 \sqrt{3})^{\mathrm{x}^2-3}=10\) Let \(t=(5+2 \sqrt{6})^{x^2-3}\) then \(\frac{1}{t}=(5-2 \sqrt{6})^{x^2-3}\) Then we have \(t+\frac{1}{t}=10\) Again \((5+2 \sqrt{6})^{x^2-3}=(5-2 \sqrt{6})\) \(=\frac{1}{5+2 \sqrt{6}}\) \(=(5+2 \sqrt{6})^{-1}\) \(\therefore x^2-3=-1\) \(\text { or } x^2=2\) \(\text { or } x= \pm \sqrt{2}\)
EAMCET-1992
Sets, Relation and Function
116960
The value of \(\left(\frac{1}{\log _3 12}+\frac{1}{\log _4 12}\right)\) is
