Question

Find the real number x satisfying \[\log_{1/2} (2x + 1 ) <-3\]?

Answer 1

this type was done by @satellite73 yesterday but am confused

Answer 2

@satellite73 can u help me out

Answer 3

yes lets change the base to make life easier

Answer 4

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

Answer 5

ok

Answer 6

if that is not clear, lets show it
by the change of base formula we know
\[\log_{\frac{1}{2}}(x)=\frac{\log_2(x)}{\log_2(\frac{1}{2})}=\frac{\log_2(x)}{-1}=-\log_2(x)\]

Answer 7

site is stuck

Answer 8

ok now we can solve
\[-\log_2(2x+1)<-3\] easily
change the signs and the sense of the inequality and get
\[\log_2(2x+1)>3\]

Answer 9

since log base 2 is an increasing function we conclude that
\[2x+1>2^3\]
\[2x+1>8\] and solve that one

Answer 10

2x+1=8 x=7/2 is it....

Answer 11

it is an inequality, not an equality

Answer 12

\(x>\frac{7}{2}\)

Answer 13

sorry

Answer 14

hope steps were clear

Answer 15

hey wat is the difference between the yesterdays question and todyss can u tell plz

Answer 16

\[ \log_{1/3} (\log_{4} (x^2-5)>0\]

Answer 17

@satellite73