Solving F ( x ) = 0.3 where F ( x ) = 1 − 200 2.5 x 2.5 Consider the function F ( x ) = 1 − 200 2.5 x 2.5 .We want to solve F ( x ) = 0.31.Since F ( x ) = 0.3 then we can say, 2.5 ln ⁡ ( 200 x ) = 0.72. ln ⁡ ( 200 x ) = 0 .283. e ln ⁡ ( 200 x ) = e .28 4. 200 x = e .28 5.Hence x = 200 e .28 6.The answer appears to be about x = 151.16Just to sum it up, the actual answer is 230.66, and this was achieved by taking the 2.5th root instead of doing the natural log. Why wasn't it the same answer?

Question

Solving   F  (  x  )  =  0.3 where   F  (  x  )  =  1  −            200              2.5                    x              2.5            Consider the function  F  (  x  )  =  1  −            200              2.5                    x              2.5              .We want to solve   F  (  x  )  =  0.31.Since   F  (  x  )  =  0.3 then we can say,   2.5  ln  ⁡  (      200    x    )  =  0.72.  ln  ⁡  (      200    x    )  =  0  .283.      e          ln      ⁡      (              200        x            )        =      e          .28      4.      200    x    =      e          .28      5.Hence   x  =      200          e              .28            6.The answer appears to be about   x  =  151.16Just to sum it up, the actual answer is   230.66, and this was achieved by taking the 2.5th root instead of doing the natural log. Why wasn&#039;t it the same answer?

Mckenna Friedman · Accepted Answer

Because it should be   ln  ⁡  0.7 on the RHS, not   0.7  F  (  x  )  =  0.3  ⇒                      200                  2.5                            x                  2.5                      =  1  −  0.3  =  0.7  ⇒    2.5  ln  ⁡            200      x        =  ln  ⁡  0.7  ⇒    ln  ⁡            200      x        =  −  0.14267  ⇒              200      x        =      e          −      0.14267        ⇒    x  =  200      e          0.14267        =                              230.67

Consider the function F(x)=1-(200^(2.5 ))/(x^{2.5)).We want to solve F(x)=0.3.

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