Maths paper 3 question

[1992]

How can u find x with this equation: it seems easy but when u start, u get stuk, well for me.
With steps shown please.

(2^x -8)^2 < 25

 

[sse2010]

is x-8 the power of 2 or only x??

 

[1992]

x is to the power of 2.

 

[MAVtKnmJ]

Expand
[(-8 + 2^x)^2 < 25]

To Get
2^(2 x)-2^(x+4)+64<25

Or
64 + 2^(2 x) - 2^(4 + x) < 25

Solve that to get

x=2 and x=3

Happy?

 

[1992]

not happy lol
Answer must be x= 1.58 and x=3.70

 

[sse2010]

its simple
expand (2^x-8)^2
dis wud bcum 2^2x-16(2^x)+64<25
=>2^2x-16(2^x)+39<0
supose 2^x=y
dis wud bcum y^2-16y+39<0
factorize it n it wud bcum 3<y<13 = 3<2^x<13
take lg=> lg3<xlg2<lg13
= lg3/lg2<x<lg13/lg2
1.58<x<3.70
hope u're happy now!

 

[MAVtKnmJ]

Well it's an inequality, you were supposed to plug in the values of x in my post into the inequality to get the correct domain/values.

Ah well, sse2010 has done it for you

P.S I might have made a minor mistake somewhere, thus our answers don't match up.

 

[1992]

ty dudes. it was tricky right?

 

[sse2010]

a bit