Question

In 2-1 2. Helen and Stephen both simplify the exponential expression e Helen: e342-7 _ eh?? _ 27 - 18 Stephen: ezh24 = 23 24 este ceila Unfortunately, they both made an error. Identify where each person made his or her error and explain in words what he or she did wrong. Then, correctly simplify the expression. Answer:

Answer 1

Some useful formulas involving exponents one should know:

Formula 1: $a^{m-n}=\frac{a^m}{a^n}$ ,for

a +0

Formula 2:
a.lnx = In(a
,for

Formula 3:
le
,for .

Formula 4:$a^{\frac{m}{n}}=(a^\frac{1}{n})^m=(a^m)^{\frac{1}{n}}$.

Now we can dig into what Helen has done wrong.

Helen's Error: $e^{\frac{4}{3}ln2-1}=\frac{e^{\frac{4}{3}ln2}}{e}=\frac{2^\frac{4}{3}}{e}$ ,up to this part she is correct.Now here she makes the error.She can't write $2^\frac{4}{3}$ as $\sqrt[4]{8}$ ,because as per the formula for exponent $a^\frac{m}{n}=(a^\frac{1}{n})^m$ ,so $2^\frac{4}{3}$ should be equal to $=(2^\frac{1}{3})^4$ ,which is not same as $(2^3)^\frac{1}{4}=8^\frac{1}{4}$ .

Stephen's error:In the first line he has written,$e^{\frac{4}{3}ln2-1}=e^{\frac{4}{3}.2-1}$,which is not correct because he is substituting as ,we know and aren't same.

So the following is the correct simplification:

$e^{\frac{4}{3}.ln 2-1}$

$=\frac{e^{\frac{4}{3}.ln2}}{e}$,(Using formula 1)

$=\frac{e^{ln2^{\frac{4}{3}}}}{e}$,(Using formula 2)

$=\frac{2^{\frac{4}{3}}}{e}$ ,(using formula 3)

$=\frac{(2^4)^{\frac{1}{3}}}{e}$,(using formula 4)

$=\frac{16^{\frac{1}{3}}}{e}$.