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hi
isn't
the same ashi bob bundy
so my problem would be:
i kinda had to edit it a million times to get it right.
hope i get the hang of it
hi,
i found the solution and it says that x shuld be in the interval (-8/5,-3/2)u(-1,0)
didn't get how they came to that solution.
so -8/5<x<-3/2 or -1<x<0
hi bobbym;
x+2 will always be the base just so you don't get confuse;
log(x+2) sqrt(5x/2+4)>1
1/2 log(x+2) (5x+8)/2 >1
log(x+2) (5x+8) - log(x+2) 2 >2
log(x+2) (5x+8) - log(x+2) 2 > log (x+2) sqr(x+2) sqr is 'square of' and the last 'x+2' is not in the base
log(x+2) (5x+8) > log(x+2) 2(sqr(x)+4x+4)
5x+8 > 2sqr(x) +8x+8
2sqr(x)<-3x
now the first case is that x is mre than 0
we divide by x
2x<-3
x<-3/2 but than x is not > 0
the second case is x<0
we divide by x s the sign changes
2x>-3
x>-3/2
x<0
so -3/2<x<0;
the third case is x=0 but then we get 0<0 which is always false
hi bobbym,
yeah that's it.
i tried something and got that -3/2<x<0, but i'm not sure.
i wanted to ask something and found this topic.
i was interested in logarithmic function and got most of it but i don't know how to do problems like this:
logx+2 sqrt(5*x/2+4)>1 where x+2 is the base.
can anyone help me?