$displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$












1












$begingroup$


Show $displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



from LS



$displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(3/2) + log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2big(frac{6(n+1)}{6}big)}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



is this right?






algebra-precalculus logarithms







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    – J. W. Tanner
    Feb 3 at 1:53
















1












$begingroup$


Show $displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



from LS



$displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(3/2) + log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2big(frac{6(n+1)}{6}big)}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



is this right?






algebra-precalculus logarithms







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  • $begingroup$
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    – J. W. Tanner
    Feb 3 at 1:53














1












1








1





$begingroup$


Show $displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



from LS



$displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(3/2) + log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2big(frac{6(n+1)}{6}big)}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



is this right?






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$




Show $displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



from LS



$displaystyle 1+ frac{log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2(3/2) + log_2(2/3(n+1))}{log_2(3/2)} = frac{log_2big(frac{6(n+1)}{6}big)}{log_2(3/2)} = frac{log_2(n+1)}{log_2(3/2)}$



is this right?






algebra-precalculus logarithms




algebra-precalculus logarithms






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share|cite|improve this question








edited Feb 2 at 20:06









El borito

664216




664216










asked Feb 2 at 19:29









Bas basBas bas

49512




49512












  • $begingroup$
    This question is similar to this question
    $endgroup$
    – J. W. Tanner
    Feb 3 at 1:53


















  • $begingroup$
    This question is similar to this question
    $endgroup$
    – J. W. Tanner
    Feb 3 at 1:53
















$begingroup$
This question is similar to this question
$endgroup$
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$begingroup$
This question is similar to this question
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– J. W. Tanner
Feb 3 at 1:53










3 Answers
3






active

oldest

votes


















0












$begingroup$

Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.



1)$$
log_{a}{frac{x}{y}}=log_{a}{left[left(frac{y}{x}right)^{-1}right]}=
-1cdotlog_{a}{frac{y}{x}}=-log_{a}{frac{y}{x}}.
$$



2)$$
log_{a}left({xy}right)=log_{a}{x}+log_{y}.
$$



$$
1+frac{log_2left[frac{2}{3}(n+1)right]}{log_2frac{3}{2}} =
1+frac{log_2frac{2}{3}+log_2left(n+1right)}{log_2frac{3}{2}} =\
1+frac{-log_2frac{3}{2}+log_2left(n+1right)}{log_2frac{3}{2}}=
1-frac{log_2frac{3}{2}}{log_2frac{3}{2}}+frac{log_2left(n+1right)}{log_2frac{3}{2}}=\
1-1+frac{log_2left(n+1right)}{log_2frac{3}{2}}=
frac{log_2left(n+1right)}{log_2frac{3}{2}}
$$




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    1












    $begingroup$

    Its right. Although for clarity, perhaps it would be better to write $displaystyle frac{log_2(2/3(n+1))}{log_2(3/2)}$ as $displaystyle frac{log_2((2/3)(n+1))}{log_2(3/2)}$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      I would write $$log_{2}frac{3}{2}+log_{2}frac{2}{3}+log_{2}(n+1)=log_{2}(n+1)$$ since $$log_{2}{frac{2}{3}}=-log_{2}{frac{3}{2}}$$






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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.



        1)$$
        log_{a}{frac{x}{y}}=log_{a}{left[left(frac{y}{x}right)^{-1}right]}=
        -1cdotlog_{a}{frac{y}{x}}=-log_{a}{frac{y}{x}}.
        $$



        2)$$
        log_{a}left({xy}right)=log_{a}{x}+log_{y}.
        $$



        $$
        1+frac{log_2left[frac{2}{3}(n+1)right]}{log_2frac{3}{2}} =
        1+frac{log_2frac{2}{3}+log_2left(n+1right)}{log_2frac{3}{2}} =\
        1+frac{-log_2frac{3}{2}+log_2left(n+1right)}{log_2frac{3}{2}}=
        1-frac{log_2frac{3}{2}}{log_2frac{3}{2}}+frac{log_2left(n+1right)}{log_2frac{3}{2}}=\
        1-1+frac{log_2left(n+1right)}{log_2frac{3}{2}}=
        frac{log_2left(n+1right)}{log_2frac{3}{2}}
        $$




        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.



          1)$$
          log_{a}{frac{x}{y}}=log_{a}{left[left(frac{y}{x}right)^{-1}right]}=
          -1cdotlog_{a}{frac{y}{x}}=-log_{a}{frac{y}{x}}.
          $$



          2)$$
          log_{a}left({xy}right)=log_{a}{x}+log_{y}.
          $$



          $$
          1+frac{log_2left[frac{2}{3}(n+1)right]}{log_2frac{3}{2}} =
          1+frac{log_2frac{2}{3}+log_2left(n+1right)}{log_2frac{3}{2}} =\
          1+frac{-log_2frac{3}{2}+log_2left(n+1right)}{log_2frac{3}{2}}=
          1-frac{log_2frac{3}{2}}{log_2frac{3}{2}}+frac{log_2left(n+1right)}{log_2frac{3}{2}}=\
          1-1+frac{log_2left(n+1right)}{log_2frac{3}{2}}=
          frac{log_2left(n+1right)}{log_2frac{3}{2}}
          $$




          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.



            1)$$
            log_{a}{frac{x}{y}}=log_{a}{left[left(frac{y}{x}right)^{-1}right]}=
            -1cdotlog_{a}{frac{y}{x}}=-log_{a}{frac{y}{x}}.
            $$



            2)$$
            log_{a}left({xy}right)=log_{a}{x}+log_{y}.
            $$



            $$
            1+frac{log_2left[frac{2}{3}(n+1)right]}{log_2frac{3}{2}} =
            1+frac{log_2frac{2}{3}+log_2left(n+1right)}{log_2frac{3}{2}} =\
            1+frac{-log_2frac{3}{2}+log_2left(n+1right)}{log_2frac{3}{2}}=
            1-frac{log_2frac{3}{2}}{log_2frac{3}{2}}+frac{log_2left(n+1right)}{log_2frac{3}{2}}=\
            1-1+frac{log_2left(n+1right)}{log_2frac{3}{2}}=
            frac{log_2left(n+1right)}{log_2frac{3}{2}}
            $$




            share|cite|improve this answer









            $endgroup$



            Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.



            1)$$
            log_{a}{frac{x}{y}}=log_{a}{left[left(frac{y}{x}right)^{-1}right]}=
            -1cdotlog_{a}{frac{y}{x}}=-log_{a}{frac{y}{x}}.
            $$



            2)$$
            log_{a}left({xy}right)=log_{a}{x}+log_{y}.
            $$



            $$
            1+frac{log_2left[frac{2}{3}(n+1)right]}{log_2frac{3}{2}} =
            1+frac{log_2frac{2}{3}+log_2left(n+1right)}{log_2frac{3}{2}} =\
            1+frac{-log_2frac{3}{2}+log_2left(n+1right)}{log_2frac{3}{2}}=
            1-frac{log_2frac{3}{2}}{log_2frac{3}{2}}+frac{log_2left(n+1right)}{log_2frac{3}{2}}=\
            1-1+frac{log_2left(n+1right)}{log_2frac{3}{2}}=
            frac{log_2left(n+1right)}{log_2frac{3}{2}}
            $$





            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 2 at 21:49









            Michael RybkinMichael Rybkin

            4,264422




            4,264422























                1












                $begingroup$

                Its right. Although for clarity, perhaps it would be better to write $displaystyle frac{log_2(2/3(n+1))}{log_2(3/2)}$ as $displaystyle frac{log_2((2/3)(n+1))}{log_2(3/2)}$






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  Its right. Although for clarity, perhaps it would be better to write $displaystyle frac{log_2(2/3(n+1))}{log_2(3/2)}$ as $displaystyle frac{log_2((2/3)(n+1))}{log_2(3/2)}$






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Its right. Although for clarity, perhaps it would be better to write $displaystyle frac{log_2(2/3(n+1))}{log_2(3/2)}$ as $displaystyle frac{log_2((2/3)(n+1))}{log_2(3/2)}$






                    share|cite|improve this answer











                    $endgroup$



                    Its right. Although for clarity, perhaps it would be better to write $displaystyle frac{log_2(2/3(n+1))}{log_2(3/2)}$ as $displaystyle frac{log_2((2/3)(n+1))}{log_2(3/2)}$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Feb 2 at 21:30









                    El borito

                    664216




                    664216










                    answered Feb 2 at 19:41









                    programmerprogrammer

                    837




                    837























                        0












                        $begingroup$

                        I would write $$log_{2}frac{3}{2}+log_{2}frac{2}{3}+log_{2}(n+1)=log_{2}(n+1)$$ since $$log_{2}{frac{2}{3}}=-log_{2}{frac{3}{2}}$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          I would write $$log_{2}frac{3}{2}+log_{2}frac{2}{3}+log_{2}(n+1)=log_{2}(n+1)$$ since $$log_{2}{frac{2}{3}}=-log_{2}{frac{3}{2}}$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            I would write $$log_{2}frac{3}{2}+log_{2}frac{2}{3}+log_{2}(n+1)=log_{2}(n+1)$$ since $$log_{2}{frac{2}{3}}=-log_{2}{frac{3}{2}}$$






                            share|cite|improve this answer









                            $endgroup$



                            I would write $$log_{2}frac{3}{2}+log_{2}frac{2}{3}+log_{2}(n+1)=log_{2}(n+1)$$ since $$log_{2}{frac{2}{3}}=-log_{2}{frac{3}{2}}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 2 at 19:39









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            79k42867




                            79k42867






























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