Taking inverse of a function that contains exponential term












0












$begingroup$


Consider we are given such a function



$$f(x) = 2^{2x-1}+3$$



How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows



Let $y = 2^{2x-1} +3$



$$ y - 3 = 2^{2x-1}$$



$$2y-6 = 2^{2x}$$



However, I dont think that is how exactly it works. Could you assist me with what I'm missing?



Regards






functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Looks good. Now take the $log()$ of both sides.
    $endgroup$
    – Michael
    Dec 18 '18 at 20:43










  • $begingroup$
    @Michael Are we supposed to take the log? I dont also know that.
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:43










  • $begingroup$
    Yea, you take the log to deal with the exponent.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:44










  • $begingroup$
    @DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:45










  • $begingroup$
    What do you mean? How would you do that?
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:45
















0












$begingroup$


Consider we are given such a function



$$f(x) = 2^{2x-1}+3$$



How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows



Let $y = 2^{2x-1} +3$



$$ y - 3 = 2^{2x-1}$$



$$2y-6 = 2^{2x}$$



However, I dont think that is how exactly it works. Could you assist me with what I'm missing?



Regards






functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Looks good. Now take the $log()$ of both sides.
    $endgroup$
    – Michael
    Dec 18 '18 at 20:43










  • $begingroup$
    @Michael Are we supposed to take the log? I dont also know that.
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:43










  • $begingroup$
    Yea, you take the log to deal with the exponent.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:44










  • $begingroup$
    @DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:45










  • $begingroup$
    What do you mean? How would you do that?
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:45














0












0








0





$begingroup$


Consider we are given such a function



$$f(x) = 2^{2x-1}+3$$



How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows



Let $y = 2^{2x-1} +3$



$$ y - 3 = 2^{2x-1}$$



$$2y-6 = 2^{2x}$$



However, I dont think that is how exactly it works. Could you assist me with what I'm missing?



Regards






functions







share|cite|improve this question









$endgroup$




Consider we are given such a function



$$f(x) = 2^{2x-1}+3$$



How does one find $f^{-1}(x)$? Here I noticed that it is a bit different to take inverse of exponential terms. My attempt is as follows



Let $y = 2^{2x-1} +3$



$$ y - 3 = 2^{2x-1}$$



$$2y-6 = 2^{2x}$$



However, I dont think that is how exactly it works. Could you assist me with what I'm missing?



Regards






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 20:40









EnzoEnzo

19917




19917








  • 2




    $begingroup$
    Looks good. Now take the $log()$ of both sides.
    $endgroup$
    – Michael
    Dec 18 '18 at 20:43










  • $begingroup$
    @Michael Are we supposed to take the log? I dont also know that.
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:43










  • $begingroup$
    Yea, you take the log to deal with the exponent.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:44










  • $begingroup$
    @DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:45










  • $begingroup$
    What do you mean? How would you do that?
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:45














  • 2




    $begingroup$
    Looks good. Now take the $log()$ of both sides.
    $endgroup$
    – Michael
    Dec 18 '18 at 20:43










  • $begingroup$
    @Michael Are we supposed to take the log? I dont also know that.
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:43










  • $begingroup$
    Yea, you take the log to deal with the exponent.
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:44










  • $begingroup$
    @DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
    $endgroup$
    – Enzo
    Dec 18 '18 at 20:45










  • $begingroup$
    What do you mean? How would you do that?
    $endgroup$
    – Don Thousand
    Dec 18 '18 at 20:45








2




2




$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43




$begingroup$
Looks good. Now take the $log()$ of both sides.
$endgroup$
– Michael
Dec 18 '18 at 20:43












$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43




$begingroup$
@Michael Are we supposed to take the log? I dont also know that.
$endgroup$
– Enzo
Dec 18 '18 at 20:43












$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44




$begingroup$
Yea, you take the log to deal with the exponent.
$endgroup$
– Don Thousand
Dec 18 '18 at 20:44












$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45




$begingroup$
@DonThousand Cannot we also just substitute $2^{2x}$ in $f(x)$?
$endgroup$
– Enzo
Dec 18 '18 at 20:45












$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45




$begingroup$
What do you mean? How would you do that?
$endgroup$
– Don Thousand
Dec 18 '18 at 20:45










1 Answer
1






active

oldest

votes


















0












$begingroup$

Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :



$$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$



Now, interchange $x leftrightarrow y$ and yield :



$$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$



Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.






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%2f3045674%2ftaking-inverse-of-a-function-that-contains-exponential-term%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









    0












    $begingroup$

    Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :



    $$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$



    Now, interchange $x leftrightarrow y$ and yield :



    $$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$



    Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :



      $$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$



      Now, interchange $x leftrightarrow y$ and yield :



      $$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$



      Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :



        $$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$



        Now, interchange $x leftrightarrow y$ and yield :



        $$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$



        Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.






        share|cite|improve this answer









        $endgroup$



        Your approach is correct. Note that $2^{2x} >0$ and thus you can always apply the $ln$ function (as mentioned-hinted in the comments). But, you also need that $2y-6 > 0 Leftrightarrow y > 3$ since you apply the $ln$ to that side as well of course. Taking this constrict into consideration, it is :



        $$ln(2y-6) = ln 2^{2x} Leftrightarrow ln(2y-6) = 2x ln 2 Leftrightarrow x = frac{ln(2y-6)}{ln 4}$$



        Now, interchange $x leftrightarrow y$ and yield :



        $$y = frac{ln(2x-6)}{ln 4} implies f^{-1}(x) = frac{ln(2x-6)}{ln 4}$$



        Don't forget that it must be $2x-6 >0 Leftrightarrow x > 3$ and $y >3 implies f(x) >3$ from the initial constrict.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 20:49









        RebellosRebellos

        14.8k31248




        14.8k31248






























            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%2f3045674%2ftaking-inverse-of-a-function-that-contains-exponential-term%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

            To store a contact into the json file from server.js file using a class in NodeJS

            Image
            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
            Read more

            Redirect URL with Chrome Remote Debugging Android Devices

            Image
            1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more