Using Logarithms − 2 n − 1 ln ⁡ 2 = − 100 ln ⁡ 10 − 100 ln ⁡ 10 = − 230 − 230 ln ⁡ ( 2 ) = − 333 − 2 n − 1 &gt; − 333 ( n − 1 ) ln ⁡ ( − 2 ) &gt; ln ⁡ ( − 333 ) Here is where I am stuck.I am not sure if this part is correct: − n − 1 = 8. Then solving we would get − 9

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Using Logarithms                    −                  2                      n            −            1                          ln        ⁡        2                            =        −        100        ln        ⁡        10                                                −        100        ln        ⁡        10                            =        −        230                                                                                    −              230                                      ln              ⁡              (              2              )                                                          =        −        333                                                −                  2                      n            −            1                                              &amp;gt;        −        333                                                (        n        −        1        )        ln        ⁡        (        −        2        )                            &amp;gt;        ln        ⁡        (        −        333        )            Here is where I am stuck.I am not sure if this part is correct:   −  n  −  1  =  8. Then solving we would get    −  9

Camron Herrera · Accepted Answer

Log is an increasing function. So you can take log in both side of an inequality and that is fine. But you can&#039;t take log of negative values like -2 The mistake is at  −      2          n      −      1        =  −  333.You missed negative sign at 333. Then if you correct the next steps it&#039;s fine. Also when you multiply both sides of an inequality by negative the sign changes.  −      2          n      −      1        l  n  2  &amp;lt;  −  100  ∗  l  n  10  −      2          n      −      1        &amp;lt;                                                  −          100          ∗          l          n          10                                                                  l          n          2                          2          n      −      1        &amp;gt;                                                  100          ∗          l          n          10                                                                  l          n          2                      n  −  1  ∗  l  n  (  2  )  &amp;gt;  l  n  (                                                  100          ∗          l          n          10                                                                  l          n          2                      )  .And this gives   n  &amp;gt;                                                  l          n          (                                                                                                                  100                  ∗                  l                  n                  10                                                                                                                                                  l                  n                  2                                                              )                                                                  l          n          2                      +  1

Using Logarithms <mtable columnalign="right left right left right left right left right left rig

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