Question

(Logarithms)
solve for x :
\[\x^{3/4(log_{2}x)^{2}+log2(x)−5/4}=sqrt{2}\]

Answer 1

@Wired help!!

Answer 2

take log both sides..and simplfy

Answer 3

and put logx=y to make a quad equation

Answer 4

find y and take its log inverse to get x

Answer 5

i was doing the same, But i am stuck at a step.

Answer 6

it will be a cubic eq

Answer 7

yes

Answer 8

i am getting a cubic eq.

Answer 9

i am not able to solve that cubic eq.

Answer 10

@himanshu31 what eq. have you got?

Answer 11

is it 3y3+4y2-g5-2log2=0

Answer 12

thats 5y

Answer 13

i got 3t^3 +4t^2 - 5t - 2 = 0

Answer 14

when u take log both side the right side will 1/2(log2)

Answer 15

hit enter wired

Answer 16

@himanshu31 what have you taken for y??

Answer 17

y=logx

Answer 18

how do u get '2'?

Answer 19

i have taken t=log(base2)x

Answer 20

same thing

Answer 21

On my RHS i have 1/log(base2)x

Answer 22

is it sqrt2 or sqrtx in question

Answer 23

sqrt2

Answer 24

@zzr0ck3r Man, I haven't done logarithms in years, give me a min lol!

In the middle of the exponent is that:
\[\log2{x} \]
or
\[\log_2{x}\]

Answer 25

im just playin man:)

Answer 26

LOL I know :)

Answer 27

log(base2)x @Wired

Answer 28

ur cubic eq is wrong ..@shubham.bagrecha

Answer 29

@himanshu31 i got this as one of my steps. \[3(\log_{2}x) ^{2}+\log_{2}x^{2}-5/2=\log_{x}2 \]

Answer 30

im to lazy to grab a pencil and join in

Answer 31

how log(base2)x^2 and how log(base x)2..???

Answer 32

for removing root, squared both sides.

Answer 33

when u take log both side the root will become 1/2 ..
there is no need to square both sides

Answer 34

i have not taken log both side.
I just used the fact that
\[a^{x}=b, x=\log_{a}b\]

Answer 35

@Wired Have you got something?

Answer 36

Still running through it. Like I said, it's been a while lol.

Answer 37

\[x ^{(3/4)(\log_{2}x)^{2} + \log_2{x}-(5/4)} = \sqrt{2}\]
\[[(3/4)(\log_{2}x)^{2} + \log_2{x}-(5/4)]log_2{x} = log_2{\sqrt{2}}\]
\[4[(3/4)(\log_{2}x)^{2} + \log_2{x}-(5/4)]log_2{x} = 4log_2{\sqrt{2}}\]
\[[3(\log_{2}x)^{2} + 4\log_2{x}-5]log_2{x} = 4log_2{\sqrt{2}}\]
\[Y=log_{2}x\]
\[(3Y^{2} + 4Y-5)Y = 4log_2{\sqrt{2}}\]
\[3Y^{3} + 4Y^{2}-5Y - 4log_2{\sqrt{2}} = 0\]
\[3Y^{3} + 4Y^{2}-5Y - 4(0.5) = 0\]
\[3Y^{3} + 4Y^{2}-5Y - 2 = 0\]

Answer 38

hey @Wired i want the value of Y
me and himanshu also got the same eq. as that of you.

Answer 39

3y^3 -3y^2 +7y^2 +2y -7y -2 =0
3y^2(y-1) + 7y(y-1) + 2(y-1)=0
try now..

Answer 40

@shubhamsrg great!!!!!
Is there any method to solve that or just trial and error??

Answer 41

i used hit and trial lol!! :P

Answer 42

1 was easy to check in this case though

Answer 43

Trial and error will kill you in a test though. There's a formula / trick, trying to find it.

Answer 44

you may use rational root theorem also (does not work everytime)

Answer 45

There's just too many formulas for it. Some are insanely huge.

Skipping that :) WolframAlpha says Y = -2, -(1/3), 1
\[Y = \log_2{x}\]
\[\log_2{x} = -2\]
\[x=2^{-2}=\frac{1}{2^{2}} = \frac{1}{4}\]


\[\log_2{x} = -\frac{1}{3}\]
\[x=2^{\frac{-1}{3}}=\frac{1}{2^{\frac{1}{3}}}=\frac{1}{\sqrt[3]{2}}\]

\[\log_2{x} = 1\]
\[x=2^{1}=2\]