How to find the roots of $f(x)= ln( frac{x+1 }{x-2})$?How to find the limit $lim limits_ {x to+infty} left [...
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How to find the roots of $f(x)= ln( frac{x+1 }{x-2})$?
How to find the limit $lim limits_ {x to+infty} left [ frac{4 ln(x+1)}{x}right]$Solving the equation $x^{ (frac{x}{123}+11)} =123$How to find the number of real roots of the given equation?Solve: $frac{1}{2}log_{frac{1}{2}}left(x-1right)>log_{frac{1}{2}}left(1-sqrt[3]{2-x}right)$How can I solve: $begin{cases} 5x - y = 2 \ 4y^2 + 4xy - 4x^2 - 1 = 0 \ end{cases}$?Find the shaded areaFind roots of $sum_i alpha_i,cos(beta_i,t)$How can I solve this hard system of equations?How to find the square roots of $z = 5-12i$How to find roots of cubic equation?
$begingroup$
I can't solve this equation:
$$lnleft(frac{x+1}{x-2}right) = 0.$$
I do:
$$begin{align*}
ln left( frac{x+1}{x-2} right)&=0\
frac{x+1}{x-2} &= 1 \
x+1&=x-2 \
x+1-x+2&=0 \
x-x+3&=0 \
3&=0
end{align*}$$
Then $x$ is?
algebra-precalculus logarithms roots
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
$endgroup$
|
show 2 more comments
$begingroup$
I can't solve this equation:
$$lnleft(frac{x+1}{x-2}right) = 0.$$
I do:
$$begin{align*}
ln left( frac{x+1}{x-2} right)&=0\
frac{x+1}{x-2} &= 1 \
x+1&=x-2 \
x+1-x+2&=0 \
x-x+3&=0 \
3&=0
end{align*}$$
Then $x$ is?
algebra-precalculus logarithms roots
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
$endgroup$
4
$begingroup$
Your steps are correct till now; the next crucial step is to conclude that there is no $x$ such that $ln (frac{x+1}{x-2}) = 0$. In other words, the equation has no solutions.
$endgroup$
– Srivatsan
Nov 14 '11 at 1:11
$begingroup$
that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24!
$endgroup$
– Totty.js
Nov 14 '11 at 1:14
$begingroup$
Plugging in $x = 2.24$ in the formula for $f(x)$ gives the value $ln (3.24 / .24) approx 2.603$, which is quite far from $0$. Are you sure you are not making some mistake in finding the zeroes in your calculator?
$endgroup$
– Srivatsan
Nov 14 '11 at 1:17
2
$begingroup$
Looking at this link: wolframalpha.com/input/?i=ln%28%28x%2B1%29%2F%28x-2%29%29 you see that at the zero of this function, there is also an imaginary part. This equation has no solution in the set of real numbers. This shows graphically why there is no solution, because when the real part is zero, then the imaginary part is nonzero!
$endgroup$
– tomcuchta
Nov 14 '11 at 1:18
$begingroup$
@Totty What do you mean by "in my calculator"? Do you mean you graphed the function? I graphed it as well and see no roots.
$endgroup$
– Austin Mohr
Nov 14 '11 at 1:19
|
show 2 more comments
$begingroup$
I can't solve this equation:
$$lnleft(frac{x+1}{x-2}right) = 0.$$
I do:
$$begin{align*}
ln left( frac{x+1}{x-2} right)&=0\
frac{x+1}{x-2} &= 1 \
x+1&=x-2 \
x+1-x+2&=0 \
x-x+3&=0 \
3&=0
end{align*}$$
Then $x$ is?
algebra-precalculus logarithms roots
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
$endgroup$
I can't solve this equation:
$$lnleft(frac{x+1}{x-2}right) = 0.$$
I do:
$$begin{align*}
ln left( frac{x+1}{x-2} right)&=0\
frac{x+1}{x-2} &= 1 \
x+1&=x-2 \
x+1-x+2&=0 \
x-x+3&=0 \
3&=0
end{align*}$$
Then $x$ is?
algebra-precalculus logarithms roots
algebra-precalculus logarithms roots
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
edited Nov 14 '11 at 2:37
Arturo Magidin
264k34590917
264k34590917
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
asked Nov 14 '11 at 1:07
Totty.jsTotty.js
1781210
1781210
4
$begingroup$
Your steps are correct till now; the next crucial step is to conclude that there is no $x$ such that $ln (frac{x+1}{x-2}) = 0$. In other words, the equation has no solutions.
$endgroup$
– Srivatsan
Nov 14 '11 at 1:11
$begingroup$
that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24!
$endgroup$
– Totty.js
Nov 14 '11 at 1:14
$begingroup$
Plugging in $x = 2.24$ in the formula for $f(x)$ gives the value $ln (3.24 / .24) approx 2.603$, which is quite far from $0$. Are you sure you are not making some mistake in finding the zeroes in your calculator?
$endgroup$
– Srivatsan
Nov 14 '11 at 1:17
2
$begingroup$
Looking at this link: wolframalpha.com/input/?i=ln%28%28x%2B1%29%2F%28x-2%29%29 you see that at the zero of this function, there is also an imaginary part. This equation has no solution in the set of real numbers. This shows graphically why there is no solution, because when the real part is zero, then the imaginary part is nonzero!
$endgroup$
– tomcuchta
Nov 14 '11 at 1:18
$begingroup$
@Totty What do you mean by "in my calculator"? Do you mean you graphed the function? I graphed it as well and see no roots.
$endgroup$
– Austin Mohr
Nov 14 '11 at 1:19
|
show 2 more comments
4
$begingroup$
Your steps are correct till now; the next crucial step is to conclude that there is no $x$ such that $ln (frac{x+1}{x-2}) = 0$. In other words, the equation has no solutions.
$endgroup$
– Srivatsan
Nov 14 '11 at 1:11
$begingroup$
that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24!
$endgroup$
– Totty.js
Nov 14 '11 at 1:14
$begingroup$
Plugging in $x = 2.24$ in the formula for $f(x)$ gives the value $ln (3.24 / .24) approx 2.603$, which is quite far from $0$. Are you sure you are not making some mistake in finding the zeroes in your calculator?
$endgroup$
– Srivatsan
Nov 14 '11 at 1:17
2
$begingroup$
Looking at this link: wolframalpha.com/input/?i=ln%28%28x%2B1%29%2F%28x-2%29%29 you see that at the zero of this function, there is also an imaginary part. This equation has no solution in the set of real numbers. This shows graphically why there is no solution, because when the real part is zero, then the imaginary part is nonzero!
$endgroup$
– tomcuchta
Nov 14 '11 at 1:18
$begingroup$
@Totty What do you mean by "in my calculator"? Do you mean you graphed the function? I graphed it as well and see no roots.
$endgroup$
– Austin Mohr
Nov 14 '11 at 1:19
4
4
$begingroup$
Your steps are correct till now; the next crucial step is to conclude that there is no $x$ such that $ln (frac{x+1}{x-2}) = 0$. In other words, the equation has no solutions.
$endgroup$
– Srivatsan
Nov 14 '11 at 1:11
$begingroup$
Your steps are correct till now; the next crucial step is to conclude that there is no $x$ such that $ln (frac{x+1}{x-2}) = 0$. In other words, the equation has no solutions.
$endgroup$
– Srivatsan
Nov 14 '11 at 1:11
$begingroup$
that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24!
$endgroup$
– Totty.js
Nov 14 '11 at 1:14
$begingroup$
that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24!
$endgroup$
– Totty.js
Nov 14 '11 at 1:14
$begingroup$
Plugging in $x = 2.24$ in the formula for $f(x)$ gives the value $ln (3.24 / .24) approx 2.603$, which is quite far from $0$. Are you sure you are not making some mistake in finding the zeroes in your calculator?
$endgroup$
– Srivatsan
Nov 14 '11 at 1:17
$begingroup$
Plugging in $x = 2.24$ in the formula for $f(x)$ gives the value $ln (3.24 / .24) approx 2.603$, which is quite far from $0$. Are you sure you are not making some mistake in finding the zeroes in your calculator?
$endgroup$
– Srivatsan
Nov 14 '11 at 1:17
2
2
$begingroup$
Looking at this link: wolframalpha.com/input/?i=ln%28%28x%2B1%29%2F%28x-2%29%29 you see that at the zero of this function, there is also an imaginary part. This equation has no solution in the set of real numbers. This shows graphically why there is no solution, because when the real part is zero, then the imaginary part is nonzero!
$endgroup$
– tomcuchta
Nov 14 '11 at 1:18
$begingroup$
Looking at this link: wolframalpha.com/input/?i=ln%28%28x%2B1%29%2F%28x-2%29%29 you see that at the zero of this function, there is also an imaginary part. This equation has no solution in the set of real numbers. This shows graphically why there is no solution, because when the real part is zero, then the imaginary part is nonzero!
$endgroup$
– tomcuchta
Nov 14 '11 at 1:18
$begingroup$
@Totty What do you mean by "in my calculator"? Do you mean you graphed the function? I graphed it as well and see no roots.
$endgroup$
– Austin Mohr
Nov 14 '11 at 1:19
$begingroup$
@Totty What do you mean by "in my calculator"? Do you mean you graphed the function? I graphed it as well and see no roots.
$endgroup$
– Austin Mohr
Nov 14 '11 at 1:19
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
What you've shown is that $frac{x+1}{x-2}$ is never equal to 1. Since 1 is the only value where natural log equals zero, the equation $log frac{x+1}{x-2} = 0 $ has no solutions.
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
$endgroup$
add a comment |
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votes
$begingroup$
What you've shown is that $frac{x+1}{x-2}$ is never equal to 1. Since 1 is the only value where natural log equals zero, the equation $log frac{x+1}{x-2} = 0 $ has no solutions.
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
$endgroup$
add a comment |
$begingroup$
What you've shown is that $frac{x+1}{x-2}$ is never equal to 1. Since 1 is the only value where natural log equals zero, the equation $log frac{x+1}{x-2} = 0 $ has no solutions.
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
$endgroup$
add a comment |
$begingroup$
What you've shown is that $frac{x+1}{x-2}$ is never equal to 1. Since 1 is the only value where natural log equals zero, the equation $log frac{x+1}{x-2} = 0 $ has no solutions.
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
$endgroup$
What you've shown is that $frac{x+1}{x-2}$ is never equal to 1. Since 1 is the only value where natural log equals zero, the equation $log frac{x+1}{x-2} = 0 $ has no solutions.
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
answered Nov 14 '11 at 1:22
Dimitrije KosticDimitrije Kostic
1,145714
1,145714
add a comment |
add a comment |
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4
$begingroup$
Your steps are correct till now; the next crucial step is to conclude that there is no $x$ such that $ln (frac{x+1}{x-2}) = 0$. In other words, the equation has no solutions.
$endgroup$
– Srivatsan
Nov 14 '11 at 1:11
$begingroup$
that's the problem! in my calculator it has zeros... on my hp48G shows up 2.24!
$endgroup$
– Totty.js
Nov 14 '11 at 1:14
$begingroup$
Plugging in $x = 2.24$ in the formula for $f(x)$ gives the value $ln (3.24 / .24) approx 2.603$, which is quite far from $0$. Are you sure you are not making some mistake in finding the zeroes in your calculator?
$endgroup$
– Srivatsan
Nov 14 '11 at 1:17
2
$begingroup$
Looking at this link: wolframalpha.com/input/?i=ln%28%28x%2B1%29%2F%28x-2%29%29 you see that at the zero of this function, there is also an imaginary part. This equation has no solution in the set of real numbers. This shows graphically why there is no solution, because when the real part is zero, then the imaginary part is nonzero!
$endgroup$
– tomcuchta
Nov 14 '11 at 1:18
$begingroup$
@Totty What do you mean by "in my calculator"? Do you mean you graphed the function? I graphed it as well and see no roots.
$endgroup$
– Austin Mohr
Nov 14 '11 at 1:19