Solve the recurrence relation by taking the logarithm of both sides and making the substitution b n = lg ⁡ a n Solve this recurrence relation: a n = ( a n − 2 a n − 1 ) 1 2 by taking the logarithm of both sides and making the substitution b n = lg ⁡ a n A couple years ago I took precalc and a couple years before that I took College Algebra to brush up since before then I'd been out of school for a couple of years already. So I'm extremely rusty and this first step, getting the logarithm of both sides, has me confused enough... Like how would I even get the logarithm of the left side?? and then what exactly is that "substitution" supposed to be substituting?So log'ing both sides I get ln ⁡ ( a n ) = 1 2 ln ⁡ ( a n − 2 ) − 1 2 ln ⁡ ( a n − 1 )

Question

Solve the recurrence relation by taking the logarithm of both sides and making the substitution       b    n    =  lg  ⁡      a    n  Solve this recurrence relation:      a    n    =            (                        a                      n            −            2                                    a                      n            −            1                              )                      1        2            by taking the logarithm of both sides and making the substitution      b    n    =  lg  ⁡      a    n  A couple years ago I took precalc and a couple years before that I took College Algebra to brush up since before then I&#039;d been out of school for a couple of years already. So I&#039;m extremely rusty and this first step, getting the logarithm of both sides, has me confused enough... Like how would I even get the logarithm of the left side?? and then what exactly is that &quot;substitution&quot; supposed to be substituting?So log&#039;ing both sides I get  ln  ⁡  (      a          n        )  =      1    2    ln  ⁡  (      a          n      −      2        )  −      1    2    ln  ⁡  (      a          n      −      1        )

MarinoneTS2 · Accepted Answer

Remember the properties of the log function:   a  log  ⁡  b  =  log  ⁡      b    a   and   log  ⁡      a    b    =  log  ⁡  a  −  log  ⁡  b. Use it to obtain the recurrence       b    n    =      1    2        b          n      −      2        −      1    2        b          n      −      1      . Then solve it using generating functions or difference equations.

DinamisGr · Accepted Answer

You have the right idea, but have not carried it far enough.You have to replace all occurrences of   ln  ⁡      a    k   with       b    k  . If you do this, from   ln  ⁡  (      a          n        )  =      1    2    ln  ⁡  (      a          n      −      2        )  −      1    2    ln  ⁡  (      a          n      −      1        ) you will get       b          n        =      1    2        b          n      −      2        −      1    2        b          n      −      1        =                    b                  n          −          2                    −              b                  n          −          1                      2  .Solve this and you can then get       a    n  .

Solve the recurrence relation by taking the logarithm of both sides and making the substitution b_n = lg a_n Solve this recurrence relation: a_n = ((a_(n-2))/(a_(n-1)))^(1/2) by taking the logarithm of both sides and making the substitution b_n = lg a_n

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