Find the range of $x$ so that $log_4{[log_{1/2}{(log_2x)}]}$'s behavior is defined












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For any $log_a b$, $a>0$, $b>0$, and $a ne 1$. How do I work my way through multiple layers of this?






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    For any $log_a b$, $a>0$, $b>0$, and $a ne 1$. How do I work my way through multiple layers of this?






    logarithms







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      $begingroup$


      For any $log_a b$, $a>0$, $b>0$, and $a ne 1$. How do I work my way through multiple layers of this?






      logarithms







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      For any $log_a b$, $a>0$, $b>0$, and $a ne 1$. How do I work my way through multiple layers of this?






      logarithms




      logarithms






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      edited Dec 12 '18 at 0:38









      N. F. Taussig

      44.1k93356




      44.1k93356










      asked Dec 12 '18 at 0:24









      Nameless KingNameless King

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          3 Answers
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          $begingroup$

          I'll assume you're talking about the "maximal" subset of the reals that makes the expression defined.



          $log_4$ only accepts values in $(0,+infty)$ so $log_frac{1}{2}$ must have its image as this set. since $log_frac{1}{2} cdot$ is identical to $-log_2 cdot$, we have that the parameters are between $2^{-0}$ and $2^{-infty} = 0$ (abuse of notation), with both $1$ and $0$ not included. Thus $log_2 x$ is between $1$ and $0$, so $x in (1, 2)$.






          share|cite|improve this answer











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          • 1




            $begingroup$
            I'm sorry but your answer is wrong. See wolframalpha.com/input/?i=log_4(log_%7B1%2F2%7D(log_2(x)))
            $endgroup$
            – Tomislav Ostojich
            Dec 12 '18 at 1:14










          • $begingroup$
            @TomislavOstojich, I got distracted. You're right; I've already fixed it. Thank you!
            $endgroup$
            – Lucas Henrique
            Dec 12 '18 at 2:07





















          2












          $begingroup$

          You need $log_{1/2}{u} > 0$ and $log_{1/2}{u} = log_2(u)/log_{2}{1/2} = -log_2{u}$. So you really need $-log_{2}{u} > 0$. Because minus signs reverse inequalities this is asking for $log_{2}{u} < 0$. Taking $f(u) = 2^u$ on both sides gives $u < 1$.



          Now $u = log_2{x}$ and the log must be non-negative for it to be a valid argument of $log_{1/2}{u}$, so $0 < log_2{x} < 1$. Applying $f(u) = 2^u$ on both sides gives $1 < x < 2$. That is the range of $x$ in which the iterated log is defined.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Just exhale and do it.



            For $log_2 x$ to exist we need $x > 0$.



            For $log_{frac 12} (log_2 x)$ so exist we need $log_2 x > 0$ so $x > 2^0 = 1$.



            For $log_4(log_{frac 12} (log_2 x))$ to exist we need $log_{frac 12} (log_2 x)> 0$ which (as $0< frac 12 < 1$) we need $log_2 x < frac 12^0=1$ so $x < 2^1=2$.



            So we need $1< x < 2$.



            Alternatively we can so $log_a^k M = frac {log_a M}{log_a a^k} = frac{log_a M}{k}$



            $log_4(log_{frac 12} (log_2 x)) = frac {log_2(log_{frac 12}(log_2 x))}{2}=frac {log_2(-log_2(log_2 x))}2$



            So $-log_2(log_2 x) < 0$ so $0< log_2 x < 1$ and so $1 < x < 2$






            share|cite|improve this answer









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              3 Answers
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              active

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              3 Answers
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              active

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              active

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              active

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              0












              $begingroup$

              I'll assume you're talking about the "maximal" subset of the reals that makes the expression defined.



              $log_4$ only accepts values in $(0,+infty)$ so $log_frac{1}{2}$ must have its image as this set. since $log_frac{1}{2} cdot$ is identical to $-log_2 cdot$, we have that the parameters are between $2^{-0}$ and $2^{-infty} = 0$ (abuse of notation), with both $1$ and $0$ not included. Thus $log_2 x$ is between $1$ and $0$, so $x in (1, 2)$.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                I'm sorry but your answer is wrong. See wolframalpha.com/input/?i=log_4(log_%7B1%2F2%7D(log_2(x)))
                $endgroup$
                – Tomislav Ostojich
                Dec 12 '18 at 1:14










              • $begingroup$
                @TomislavOstojich, I got distracted. You're right; I've already fixed it. Thank you!
                $endgroup$
                – Lucas Henrique
                Dec 12 '18 at 2:07


















              0












              $begingroup$

              I'll assume you're talking about the "maximal" subset of the reals that makes the expression defined.



              $log_4$ only accepts values in $(0,+infty)$ so $log_frac{1}{2}$ must have its image as this set. since $log_frac{1}{2} cdot$ is identical to $-log_2 cdot$, we have that the parameters are between $2^{-0}$ and $2^{-infty} = 0$ (abuse of notation), with both $1$ and $0$ not included. Thus $log_2 x$ is between $1$ and $0$, so $x in (1, 2)$.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                I'm sorry but your answer is wrong. See wolframalpha.com/input/?i=log_4(log_%7B1%2F2%7D(log_2(x)))
                $endgroup$
                – Tomislav Ostojich
                Dec 12 '18 at 1:14










              • $begingroup$
                @TomislavOstojich, I got distracted. You're right; I've already fixed it. Thank you!
                $endgroup$
                – Lucas Henrique
                Dec 12 '18 at 2:07
















              0












              0








              0





              $begingroup$

              I'll assume you're talking about the "maximal" subset of the reals that makes the expression defined.



              $log_4$ only accepts values in $(0,+infty)$ so $log_frac{1}{2}$ must have its image as this set. since $log_frac{1}{2} cdot$ is identical to $-log_2 cdot$, we have that the parameters are between $2^{-0}$ and $2^{-infty} = 0$ (abuse of notation), with both $1$ and $0$ not included. Thus $log_2 x$ is between $1$ and $0$, so $x in (1, 2)$.






              share|cite|improve this answer











              $endgroup$



              I'll assume you're talking about the "maximal" subset of the reals that makes the expression defined.



              $log_4$ only accepts values in $(0,+infty)$ so $log_frac{1}{2}$ must have its image as this set. since $log_frac{1}{2} cdot$ is identical to $-log_2 cdot$, we have that the parameters are between $2^{-0}$ and $2^{-infty} = 0$ (abuse of notation), with both $1$ and $0$ not included. Thus $log_2 x$ is between $1$ and $0$, so $x in (1, 2)$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 12 '18 at 1:41

























              answered Dec 12 '18 at 1:10









              Lucas HenriqueLucas Henrique

              1,039414




              1,039414








              • 1




                $begingroup$
                I'm sorry but your answer is wrong. See wolframalpha.com/input/?i=log_4(log_%7B1%2F2%7D(log_2(x)))
                $endgroup$
                – Tomislav Ostojich
                Dec 12 '18 at 1:14










              • $begingroup$
                @TomislavOstojich, I got distracted. You're right; I've already fixed it. Thank you!
                $endgroup$
                – Lucas Henrique
                Dec 12 '18 at 2:07
















              • 1




                $begingroup$
                I'm sorry but your answer is wrong. See wolframalpha.com/input/?i=log_4(log_%7B1%2F2%7D(log_2(x)))
                $endgroup$
                – Tomislav Ostojich
                Dec 12 '18 at 1:14










              • $begingroup$
                @TomislavOstojich, I got distracted. You're right; I've already fixed it. Thank you!
                $endgroup$
                – Lucas Henrique
                Dec 12 '18 at 2:07










              1




              1




              $begingroup$
              I'm sorry but your answer is wrong. See wolframalpha.com/input/?i=log_4(log_%7B1%2F2%7D(log_2(x)))
              $endgroup$
              – Tomislav Ostojich
              Dec 12 '18 at 1:14




              $begingroup$
              I'm sorry but your answer is wrong. See wolframalpha.com/input/?i=log_4(log_%7B1%2F2%7D(log_2(x)))
              $endgroup$
              – Tomislav Ostojich
              Dec 12 '18 at 1:14












              $begingroup$
              @TomislavOstojich, I got distracted. You're right; I've already fixed it. Thank you!
              $endgroup$
              – Lucas Henrique
              Dec 12 '18 at 2:07






              $begingroup$
              @TomislavOstojich, I got distracted. You're right; I've already fixed it. Thank you!
              $endgroup$
              – Lucas Henrique
              Dec 12 '18 at 2:07













              2












              $begingroup$

              You need $log_{1/2}{u} > 0$ and $log_{1/2}{u} = log_2(u)/log_{2}{1/2} = -log_2{u}$. So you really need $-log_{2}{u} > 0$. Because minus signs reverse inequalities this is asking for $log_{2}{u} < 0$. Taking $f(u) = 2^u$ on both sides gives $u < 1$.



              Now $u = log_2{x}$ and the log must be non-negative for it to be a valid argument of $log_{1/2}{u}$, so $0 < log_2{x} < 1$. Applying $f(u) = 2^u$ on both sides gives $1 < x < 2$. That is the range of $x$ in which the iterated log is defined.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                You need $log_{1/2}{u} > 0$ and $log_{1/2}{u} = log_2(u)/log_{2}{1/2} = -log_2{u}$. So you really need $-log_{2}{u} > 0$. Because minus signs reverse inequalities this is asking for $log_{2}{u} < 0$. Taking $f(u) = 2^u$ on both sides gives $u < 1$.



                Now $u = log_2{x}$ and the log must be non-negative for it to be a valid argument of $log_{1/2}{u}$, so $0 < log_2{x} < 1$. Applying $f(u) = 2^u$ on both sides gives $1 < x < 2$. That is the range of $x$ in which the iterated log is defined.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You need $log_{1/2}{u} > 0$ and $log_{1/2}{u} = log_2(u)/log_{2}{1/2} = -log_2{u}$. So you really need $-log_{2}{u} > 0$. Because minus signs reverse inequalities this is asking for $log_{2}{u} < 0$. Taking $f(u) = 2^u$ on both sides gives $u < 1$.



                  Now $u = log_2{x}$ and the log must be non-negative for it to be a valid argument of $log_{1/2}{u}$, so $0 < log_2{x} < 1$. Applying $f(u) = 2^u$ on both sides gives $1 < x < 2$. That is the range of $x$ in which the iterated log is defined.






                  share|cite|improve this answer











                  $endgroup$



                  You need $log_{1/2}{u} > 0$ and $log_{1/2}{u} = log_2(u)/log_{2}{1/2} = -log_2{u}$. So you really need $-log_{2}{u} > 0$. Because minus signs reverse inequalities this is asking for $log_{2}{u} < 0$. Taking $f(u) = 2^u$ on both sides gives $u < 1$.



                  Now $u = log_2{x}$ and the log must be non-negative for it to be a valid argument of $log_{1/2}{u}$, so $0 < log_2{x} < 1$. Applying $f(u) = 2^u$ on both sides gives $1 < x < 2$. That is the range of $x$ in which the iterated log is defined.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 12 '18 at 1:17

























                  answered Dec 12 '18 at 0:48









                  Tomislav OstojichTomislav Ostojich

                  508414




                  508414























                      1












                      $begingroup$

                      Just exhale and do it.



                      For $log_2 x$ to exist we need $x > 0$.



                      For $log_{frac 12} (log_2 x)$ so exist we need $log_2 x > 0$ so $x > 2^0 = 1$.



                      For $log_4(log_{frac 12} (log_2 x))$ to exist we need $log_{frac 12} (log_2 x)> 0$ which (as $0< frac 12 < 1$) we need $log_2 x < frac 12^0=1$ so $x < 2^1=2$.



                      So we need $1< x < 2$.



                      Alternatively we can so $log_a^k M = frac {log_a M}{log_a a^k} = frac{log_a M}{k}$



                      $log_4(log_{frac 12} (log_2 x)) = frac {log_2(log_{frac 12}(log_2 x))}{2}=frac {log_2(-log_2(log_2 x))}2$



                      So $-log_2(log_2 x) < 0$ so $0< log_2 x < 1$ and so $1 < x < 2$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Just exhale and do it.



                        For $log_2 x$ to exist we need $x > 0$.



                        For $log_{frac 12} (log_2 x)$ so exist we need $log_2 x > 0$ so $x > 2^0 = 1$.



                        For $log_4(log_{frac 12} (log_2 x))$ to exist we need $log_{frac 12} (log_2 x)> 0$ which (as $0< frac 12 < 1$) we need $log_2 x < frac 12^0=1$ so $x < 2^1=2$.



                        So we need $1< x < 2$.



                        Alternatively we can so $log_a^k M = frac {log_a M}{log_a a^k} = frac{log_a M}{k}$



                        $log_4(log_{frac 12} (log_2 x)) = frac {log_2(log_{frac 12}(log_2 x))}{2}=frac {log_2(-log_2(log_2 x))}2$



                        So $-log_2(log_2 x) < 0$ so $0< log_2 x < 1$ and so $1 < x < 2$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Just exhale and do it.



                          For $log_2 x$ to exist we need $x > 0$.



                          For $log_{frac 12} (log_2 x)$ so exist we need $log_2 x > 0$ so $x > 2^0 = 1$.



                          For $log_4(log_{frac 12} (log_2 x))$ to exist we need $log_{frac 12} (log_2 x)> 0$ which (as $0< frac 12 < 1$) we need $log_2 x < frac 12^0=1$ so $x < 2^1=2$.



                          So we need $1< x < 2$.



                          Alternatively we can so $log_a^k M = frac {log_a M}{log_a a^k} = frac{log_a M}{k}$



                          $log_4(log_{frac 12} (log_2 x)) = frac {log_2(log_{frac 12}(log_2 x))}{2}=frac {log_2(-log_2(log_2 x))}2$



                          So $-log_2(log_2 x) < 0$ so $0< log_2 x < 1$ and so $1 < x < 2$






                          share|cite|improve this answer









                          $endgroup$



                          Just exhale and do it.



                          For $log_2 x$ to exist we need $x > 0$.



                          For $log_{frac 12} (log_2 x)$ so exist we need $log_2 x > 0$ so $x > 2^0 = 1$.



                          For $log_4(log_{frac 12} (log_2 x))$ to exist we need $log_{frac 12} (log_2 x)> 0$ which (as $0< frac 12 < 1$) we need $log_2 x < frac 12^0=1$ so $x < 2^1=2$.



                          So we need $1< x < 2$.



                          Alternatively we can so $log_a^k M = frac {log_a M}{log_a a^k} = frac{log_a M}{k}$



                          $log_4(log_{frac 12} (log_2 x)) = frac {log_2(log_{frac 12}(log_2 x))}{2}=frac {log_2(-log_2(log_2 x))}2$



                          So $-log_2(log_2 x) < 0$ so $0< log_2 x < 1$ and so $1 < x < 2$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 12 '18 at 15:31









                          fleabloodfleablood

                          69.8k22685




                          69.8k22685






























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