Question

0.6^x-1/9>-5/18

But the issue is how do I get
s= [-4;+infinite)
s= [-1/4;+infinite)
0r s= [1/4;+infinite)

but i keep gettign 26/100 or rather 0.26

Answer 1

\[\frac{ 0.6^x -1 }{ 9} > -\frac{ 5 }{ 18 }\]
multiply by 9
\[0.6^x -1> -\frac{ 5 }{ 18 }\times 9\]
\[0.6^x-1>-\frac{5}{2}\]
\[0.6^x > -\frac{ 5 }{ 2 }+1\]
\[0.6^x>-\frac{ 3 }{ 2 }>-1.5 \]
\[x log(0.6)>log(-1.5)\]
have i written the correct statement?

Answer 2

domain of log is >=0

Answer 3

well the original is 0.6^X -1/9 > -5/18 but what u did is so close to it that i might just be the right solution

Answer 4

btw it 0.6666666666666666666666666666666666666666666666666666666666666666

Answer 5

let x=0.66666...
10x=6.66666...
subtract
9x=6
x=6/9=2/3
\[\frac{( \frac{ 2 }{ 3 })^x-1 }{ 9}>-\frac{ 5 }{ 18 }\]
\[(\frac{ 2 }{ 3 })^x-1=-\frac{ 5 }{ 18 }\times 9\]
\[(\frac{ 2 }{ 3 })^x>-\frac{ 5 }{ 2 }+1\]
\[(\frac{ 2 }{ 3 })^x>-\frac{ 3 }{ 2 }\]
\[(\frac{ 3 }{ 2 })^{-x}>-\frac{ 3 }{ 2 }\]
for all values of x L.H.S. is positive which is >-3/2
solution is all real valuues of x.