Question

The number of integers satisfying the inequality is-
\[\log_{2}\sqrt{x} - (\log_{1/4}x^{2})^{2} + 1 > 0\]

Ans.3

Answer 1

Note :
\[(\log_{1/4}x^{2})^{2} = \log^{2}_{1/4}x^{2}\]

Answer 2

pls anyone give me solution.

Answer 3

@lgbasallote Can you help??

Answer 4

no idea sorry

Answer 5

ok thanks

Answer 6

@nitz Now,Only you could help me on OS.

Answer 7

are you sure the answer is 3
because i am getting 2
i simplified the equation by changing the base to natural log
then i got a quadratic equation in log with inequality solving it i got
2.429>x>0.581

Answer 8

yes it is 3.

Answer 9

@nitz

Answer 10

@nitz Help i am coming back after 30mins.

Answer 11

ok

Answer 12

@satellite73 THE BOSS!

Answer 13

SHUBHAM THE SOLUTION IS GOING TOO LENGHTY

Answer 14

@nitz tell me starting steps

Answer 15

according to me
\[\log_{2}\sqrt{x}=(1/2)\log_{2} (x)\]

Answer 16

\[(\log_{1/4} (x ^{2}))^{2}=(2\log_{1/4}x)^{2} \]

Answer 17

wait

Answer 18

explain the above step as to how we arrived at it?

Answer 19

which one ?

Answer 20

\[\log_{b}(x ^{n})=n \log_{b} (x)\]

Answer 21

the last one

Answer 22

ok

Answer 23

where is 2?

Answer 24

not 2, where is 1 from the ques.?

Answer 25

1 from question????????

Answer 26

yes

Answer 27

matlab kya?

Answer 28

guess we want to write everything with one base right?

Answer 29

yes

Answer 30

मतलब ques. में जो 1 हे उसका क्या हुआ ?

Answer 31

@nitz i got the eq, x+2x+2>0

Answer 32

final one

Answer 33

wait

Answer 34

i am getting a quadratic in final answer

Answer 35

yes me too
it is - 2x^2 +x +1>0

Answer 36

that is 2x^2

Answer 37

without negative

Answer 38

\[t-2t ^{2}+2>0\]

Answer 39

i got \(2x^2+x-1<0\)

Answer 40

minus plus kab bana???

Answer 41

wait checking

Answer 42

assuming of course that \(x=\log_{\frac{1}{4}}(x)\)

Answer 43

i am assuming t=\[\log_{2}(x) \]

Answer 44

but i am algebra challenged today so it could be wrong

Answer 45

hello james!

Answer 46

Hi sat!
*************

If
\[\log_{2}\sqrt{x} - (\log_{1/4}x^{2})^{2} + 1 > 0\]
then writing everything with base e,
\[ \frac{(1/2) \ln x}{\ln 2} - \left(\frac{2\ln x}{-2 \ln 2}\right)^2 + 1 > 0 \]
Now simplify this one or two more steps.

Answer 47

i used \(\log_2(x)=-2\log_{\frac{1}{4}}(x)\)

Answer 48

Makes sense.

Answer 49

ya i too got same as james

Answer 50

works out nicely because you get
\[-t-2t^2+1>0\]
\[2t^2+t-1<0\]
\[(2t-1)(t+1)<0\]
\[-1<t<\frac{1}{2}\]

Answer 51

now i am getting \[2t^{2}+2>0\]

Answer 52

where \(t=\log_{\frac{1}{4}}(x)\)

Answer 53

@satellite73 what values you getting after replacing t?

Answer 54

you mean after \(-1<\log_{\frac{1}{4}}(x)<\frac{1}{2}\)?

Answer 55

yes

Answer 56

i am confuse with powers.

Answer 57

ok well the base here is \(\frac{1}{4}\) so this log is a strictly decreasing function
we solve for \(x\) via
\[(\frac{1}{4})^{-1}>x>(\frac{1}{4})^{\frac{1}{2}}\]

Answer 58

@satellite73 your ans. is correct.
Now steps .

Answer 59

or in other words
\[\frac{\sqrt{2}}{2}<x<4\]

Answer 60

1/2 < x < 4

Answer 61

yeah that

Answer 62

now tell me steps from starting

Answer 63

so i guess integer solutions are 1, 2 and 3 halas

Answer 64

you want all the gory details?

Answer 65

yes

Answer 66

first i wrote everything in terms of \(\log_{\frac{1}{4}}(x)\)

Answer 67

because that is what is being squared
so lets start from the beginning

Answer 68

what eq. you got after writing in terms of base 1/4

Answer 69

\[\log_2(\sqrt{x})=\frac{1}{2}\log_2(x)=\frac{1}{2}\times (-2\log_{\frac{1}{4}}(x))=-\log_{\frac{1}{4}}(x)\]

Answer 70

\[\log_{\frac{1}{4}}(x^2)=2\log_{\frac{1}{4}}(x)\]

Answer 71

replacing \(t=\log_{\frac{1}{4}}(x)\) you get
\[-t-4t^2+1>0\] damn i forgot to square the two first time. ok lets see
\[4t^2+t-1<0\] now my pretty solution is all messed up

Answer 72

damn damn damn
sam? any ideas?

Answer 73

nowonder , I'm getting -t-4t^2+1>0 just now :)

Answer 74

oh hell i thought this was so nice
now it is a mess

Answer 75

ok,
-t-4t^2+1>0
quadratic equation
\[t=\frac{-(1)\pm \sqrt{1-4(-4)}}{2(-4)}\]
\[t=\frac{1\pm \sqrt{17}}{8}\]

Answer 76

lets change to \(\log_2(x)\)

Answer 77

I tried changing to log base 2, but doesn't work out
\[4\log_2 \sqrt{x}-(\log_2x^2)^2+4=0\]

Answer 78

\[\frac{1}{2}\log_2(x)-(-\log_2(x))^2+1>0\]

Answer 79

damn damn damn

Answer 80

ok same thing either way

Answer 81

well i messed this all up now i am wondering what to do to get the answer

Answer 82

now i am only getting two integer solutions, 1 and 2

Answer 83

log(sub 2)(sqrt(x))-(log(sub 1/4)(x^2))^2+1>0

ln(sqrt(x))/ln(2)-(ln(x^2)/ln(1/4))^2+1>0

ln((x)^1/2)/ln(2)-(2ln(x)/ln(1/4))^2+1>0

ln(x)/ln(4)-4ln^2(x)/ln(1/4)^2+1>0

ln(x)/ln(4)-4ln^2(x)/ln(1/4)^2+1>0

ln(x)=t

At+Bt^2+C>0

A=1/ln(4)=0.72
B=-4/ln(1/4)^2=-2.08
C=1

t=0.88772=ln(x)=2.429
t=-0.5415=ln(x)=0.581

What about this?

Answer 84

I got it,
from
\[t=\frac{-1\pm \sqrt{17}}{8}\]
\[\log_{1/4}x=\frac{-1\pm \sqrt{17}}{8}\]
\[\huge x=\frac{1}{4}^{\frac{(-1\pm \sqrt{17})}{8}}\]
------------------------

Answer 85

then................

Answer 86

I missed a minus sign in the exponent just now, lol

Answer 87

@shubham.bagrecha that's it

Answer 88

yeah that is it, but with this weird answer there are only two solutions
i am going to make a guess that there was an error somewhere in the problem. maybe it should not have been \(\log_{\frac{1}{4}}(x^2)\) but rather \(\log_{\frac{1}{4}}(x)\) which would make the solution much nicer and in fact the three integer solutions would be 1, 2, 3

Answer 89

finally i got the answer.