Mixing
Find range of $x$ if $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$
$begingroup$
If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______
My Attempt
$$
log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
frac{4(3x+1)}{x^2+10x}leq1\
implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
$$
My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?
inequality logarithms
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
$endgroup$
add a comment |
$begingroup$
If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______
My Attempt
$$
log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
frac{4(3x+1)}{x^2+10x}leq1\
implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
$$
My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?
inequality logarithms
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
$endgroup$
add a comment |
$begingroup$
If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______
My Attempt
$$
log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
frac{4(3x+1)}{x^2+10x}leq1\
implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
$$
My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?
inequality logarithms
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
$endgroup$
If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______
My Attempt
$$
log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
frac{4(3x+1)}{x^2+10x}leq1\
implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
$$
My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?
inequality logarithms
inequality logarithms
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
edited Jan 23 at 19:14
ss1729
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
asked Jan 23 at 14:21
ss1729ss1729
2,00311024
2,00311024
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$
is true!
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
$begingroup$
ohh i can't believe i did not see the domain of log got to be greater than zero.
$endgroup$
– ss1729
Jan 23 at 19:18
add a comment |
$begingroup$
By your work we need to solve
$$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$
The first by the interval's method gives
$$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
$$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
$$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
$$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
$$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$
edited Jan 23 at 14:51
Bernard
122k741116
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$
is true!
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
$begingroup$
ohh i can't believe i did not see the domain of log got to be greater than zero.
$endgroup$
– ss1729
Jan 23 at 19:18
add a comment |
$begingroup$
You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$
is true!
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
$begingroup$
ohh i can't believe i did not see the domain of log got to be greater than zero.
$endgroup$
– ss1729
Jan 23 at 19:18
add a comment |
$begingroup$
You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$
is true!
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$
is true!
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
answered Jan 23 at 14:27
Maria MazurMaria Mazur
46.7k1260120
46.7k1260120
$begingroup$
ohh i can't believe i did not see the domain of log got to be greater than zero.
$endgroup$
– ss1729
Jan 23 at 19:18
add a comment |
$begingroup$
ohh i can't believe i did not see the domain of log got to be greater than zero.
$endgroup$
– ss1729
Jan 23 at 19:18
$begingroup$
ohh i can't believe i did not see the domain of log got to be greater than zero.
$endgroup$
– ss1729
Jan 23 at 19:18
$begingroup$
ohh i can't believe i did not see the domain of log got to be greater than zero.
$endgroup$
– ss1729
Jan 23 at 19:18
add a comment |
$begingroup$
By your work we need to solve
$$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$
The first by the interval's method gives
$$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
$$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By your work we need to solve
$$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$
The first by the interval's method gives
$$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
$$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
By your work we need to solve
$$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$
The first by the interval's method gives
$$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
$$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
By your work we need to solve
$$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$
The first by the interval's method gives
$$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
$$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 23 at 14:36
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 14:27
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
$begingroup$
Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
$$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
$$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
$$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$
edited Jan 23 at 14:51
Bernard
122k741116
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
$$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
$$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
$$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$
edited Jan 23 at 14:51
Bernard
122k741116
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
$$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
$$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
$$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$
edited Jan 23 at 14:51
Bernard
122k741116
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
$$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
$$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
$$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$
edited Jan 23 at 14:51
Bernard
122k741116
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
edited Jan 23 at 14:51
Bernard
122k741116
edited Jan 23 at 14:51
Bernard
122k741116
edited Jan 23 at 14:51
Bernard
122k741116
122k741116
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered Jan 23 at 14:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
77.7k42866
add a comment |
add a comment |
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