Question

solve for x :
4^[x-((root(x^2) - 5)) - 12*2^x-1-((root(x^2)-5)) + 8 = 0
if confused see attachment.

Answer 1

Your square bracket doesn't have a closing one, so there's no way to solve it without you posting where the bracket goes.

Answer 2

wait i am attaching a file.

Answer 3

Is that supposed to be like this?
\[4^{x-\sqrt{x^{2}-5}} - 12^{x-1-\sqrt{x^{2}-5}} -+8 = 0\]

Answer 4

yes

Answer 5

what have you typed in equation??
i was not able to make that equation.

Answer 6

4^{x-sqrt{x^{2}-5}}-12^{x-1-sqrt{x^{2}-5}}+8=0

Answer 7

oh thanks .
now help me with my question.

Answer 8

\[4^{x-\sqrt{x^{2}-5}}-12^{x-1-\sqrt{x^{2}-5}}+8=0\]
\[4^{1}*4^{x-\sqrt{x^{2}-5}-1}-12^{x-1-\sqrt{x^{2}-5}}+8=0\]
\[y = x-\sqrt{x^{2}-5}-1\]

\[4^{1}*4^{y}-12^{y}+8=0\]

Solve for Y from there, then replace y with the formula above and solve for x.

Answer 9

oh i am sorry
in the question there is 12*2..............................

Answer 10

\[4^{x-\sqrt{x^{2}-5}}-12*2^{x-1-\sqrt{x^{2}-5}}+8=0\]

Answer 11

Oh well hell, that changes everything lol

Answer 12

Do you get the idea though from what I posted earlier?

Answer 13

i know that idea but still i am not getting the answer.

Answer 14

Keep in mind that:
\[b^{S+T} = b^{S}*b^{T}\]
\[(b^{S})^{T} = b^{S*T}\]
\[(b*c)^{S} = b^{S}*c^{S}\]

Answer 15

shubham is x=3 and x=9/4 your answer??

Answer 16

All you're doing is using those rules over and over again. That's why simplifying the exponents makes it easier to manipulate. When you're not looking at all of that junk it is easier.

You can start out with a simpler simplification (I know, sounds odd):
\[4^{x-\sqrt{x^{2}-5}}-12*2^{x-1-\sqrt{x^{2}-5}}+8=0\]

\[y = \sqrt{x^{2}-5}\]

Write out the equation with that substitution.

Answer 17

yes or no?

Answer 18

yes nitz, that's correct!!!!!

Answer 19

should i explain???

Answer 20

yes

Answer 21

\[4^{x-\sqrt{^{x ^{2-}5}}-1+1}\]

Answer 22

now \[4^{1}.4^{(x-1)-\sqrt(x ^{2}-5}\]

Answer 23

put\[2^{(x-1)-\sqrt{x ^{2}-5}}=t\]

Answer 24

so the equation becomes
4t^2-12t+8=0

Answer 25

ie t^2-3t+2=0
giving t=1 and t=2

Answer 26

got till here????

Answer 27

got d answer.

Answer 28

thanks both of you.

Answer 29

\[2^{(x-1)-\sqrt{x ^{2}-5}}=1\]
and \[2^{(x-1)-\sqrt{x ^{2}-5}}=2\]

Answer 30

\[4^{x-y}-12*2^{x-1-y}+8=0\]
\[4^{1}*4^{x-y-1}-12*2^{x-1-y}+8=0\]
\[4^{1}*4^{x-y-1}-12*2^{x-y-1}+8=0\]
\[4^{1}*(2*2)^{x-y-1}-12*2^{x-y-1}+8=0\]
\[4^*(2)^{x-y-1}*(2)^{x-y-1}-12*2^{x-y-1}+8=0\]
\[z = (2)^{x-y-1}\]
\[4z^{2}-12z+8=0\]

Answer 31

Same thing lol

Answer 32

so i think rest you can sol;veeeee

Answer 33

thanks.