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 [...

Use Mercury as quenching liquid for swords?

Help! My Character is too much for her story!

I am the person who abides by rules but breaks the rules . Who am I

How to recover against Snake as a heavyweight character?

Having the player face themselves after the mid-game

Create chunks from an array

Has a sovereign Communist government ever run, and conceded loss, on a fair election?

Iron deposits mined from under the city

Why is my explanation wrong?

Was it really inappropriate to write a pull request for the company I interviewed with?

Book where society has been split into 2 with a wall down the middle where one side embraced high tech whereas other side were totally against tech

Vector-transposing function

Trigger on Custom Object Share

A running toilet that stops itself

Other authors are notified except me, good or bad sign?

ESPP--any reason not to go all in?

Why are partial PostgreSQL HASH indices not smaller than full indices?

Can I frame a new window without adding jack studs?

How do DNA viruses keep themselves in the nucleus without inserting themselves into genome?

Why is there an extra space when I type "ls" on the Desktop?

How do I say in Japanese "I'm doing much better/healing well"?

Too soon for a plot twist?

Precision notation for voltmeters

Why does a car's steering wheel get lighter with increasing speed



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?













3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














3












3








3





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 14 '11 at 2:37









Arturo Magidin

264k34590917




264k34590917










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














  • 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










1 Answer
1






active

oldest

votes


















8












$begingroup$

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.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f81841%2fhow-to-find-the-roots-of-fx-ln-fracx1-x-2%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 14 '11 at 1:22









        Dimitrije KosticDimitrije Kostic

        1,145714




        1,145714






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f81841%2fhow-to-find-the-roots-of-fx-ln-fracx1-x-2%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Nidaros erkebispedøme

            Image
            For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bj...
            Read more

            Birsay

            Image
            Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser d...
            Read more

            Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...

            Image
            Do VLANs within a subnet need to have their own subnet for router on a stick? What are the differences between the usage of 'it' and 'they'? Dragon forelimb placement How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative? What typically incentivizes a professor to change jobs to a lower ranking university? How to write a macro that is braces sensitive? Modeling an IPv4 Address Writing rule stating superpower from different root cause is bad writing Can I ask the recruiters in my resume to put the reason why I am rejected? Approximately how much travel t...
            Read more