Logarithm rules i.e. $-1/2 cdot log_2(2/9)$ to $2/9cdot log_2(2)$?












1












$begingroup$


I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48
















1












$begingroup$


I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48














1












1








1





$begingroup$


I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$




I'm struggling to understand the flow of calculation as shown in the picture below.



enter image description here



It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.



Thank you! :)






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 21:21









callculus

17.9k31427




17.9k31427










asked Jan 7 at 20:28









Giga2001Giga2001

103




103








  • 2




    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48














  • 2




    $begingroup$
    While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
    $endgroup$
    – Matteo
    Jan 7 at 20:46








  • 3




    $begingroup$
    The substitution isn’t correct.
    $endgroup$
    – KM101
    Jan 7 at 20:48








2




2




$begingroup$
While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
$endgroup$
– Matteo
Jan 7 at 20:46






$begingroup$
While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-frac{1}{2} log_2 left(frac{2}{9}right) neq frac{2}{9}log_2 2 = frac{2]{9}$.
$endgroup$
– Matteo
Jan 7 at 20:46






3




3




$begingroup$
The substitution isn’t correct.
$endgroup$
– KM101
Jan 7 at 20:48




$begingroup$
The substitution isn’t correct.
$endgroup$
– KM101
Jan 7 at 20:48










1 Answer
1






active

oldest

votes


















0












$begingroup$

I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



begin{align}
&1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
&textrm{Denominator: You're just adding the same thing 9 times}\\
= &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
= &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
end{align}



For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






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%2f3065459%2flogarithm-rules-i-e-1-2-cdot-log-22-9-to-2-9-cdot-log-22%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












    $begingroup$

    I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



    begin{align}
    &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
    &textrm{Denominator: You're just adding the same thing 9 times}\\
    = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
    = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
    end{align}



    For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



      begin{align}
      &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
      &textrm{Denominator: You're just adding the same thing 9 times}\\
      = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
      = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
      end{align}



      For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



        begin{align}
        &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
        &textrm{Denominator: You're just adding the same thing 9 times}\\
        = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
        end{align}



        For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.






        share|cite|improve this answer









        $endgroup$



        I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.



        begin{align}
        &1 - frac{-frac{1}{3}log_2left(frac{1}{3}right) - frac{1}{2}log_2left(frac{2}{9}right)}{-sumlimits_{s = 1}^9 frac{1}{9}log_2left(frac{1}{9}right)}\\
        &textrm{Denominator: You're just adding the same thing 9 times}\\
        = &1 - frac{frac{1}{3}log_2left(left(frac{1}{3}right)^{-1}right) + frac{1}{2}log_2left(left(frac{2}{9}right)^{-1}right)}{-9 cdot frac{1}{9}log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{-log_2left(frac{1}{9}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(left(frac{1}{9}right)^{-1}right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(frac{9}{2}right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}left(log_2left(9right) - log_2left(2right)right)}{log_2left(9right)}\\
        = &1 - frac{frac{1}{3}log_2left(3right) + frac{1}{2}log_2left(9right) - frac{1}{2}}{log_2left(9right)}\\
        end{align}



        For this to be equal to the second step, you'd need $frac{1}{2}log_2left(9right) - frac{1}{2} = frac{2}{9}log_2left(2right) = frac{2}{9}$ to be true. It's not: $frac{1}{2}log_2left(9right) - frac{1}{2} approx 1.1$ and $frac{2}{9} approx 0.2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 3:46









        PiKindOfGuyPiKindOfGuy

        18611




        18611






























            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%2f3065459%2flogarithm-rules-i-e-1-2-cdot-log-22-9-to-2-9-cdot-log-22%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

            'app-layout' is not a known element: how to share Component with different Modules

            Image
            0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
            Read more

            android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
            Read more

            WPF add header to Image with URL pettitions [duplicate]

            Image
            up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
            Read more