In Wolfram Alpha this statement is false. But how? Because $(a)^{\frac{b}{c}}=(a^b)^{\frac{1}{c}}$. Is there any condition. Please tell me.
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3See the case of Negative numbers with rational exponents – Mauro ALLEGRANZA Oct 05 '20 at 11:10
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$a^{xy}=(a^x)^y$ doesn't always work for $a<0$. – Gerry Myerson Oct 05 '20 at 12:02
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2"Because $(a)^{\frac{b}{c}}=(a^b)^{\frac{1}{c}}$." Whoever or wherever you think you learned this from did not emphasize that it works only for positive $a$ strongly enough. Consider also $-1=(-1)^1 = (-1)^{2/2} \color{red}{=}((-1)^2)^{1/2} = 1^{1/2}=1$ – JMoravitz Oct 05 '20 at 12:04
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Look here !!
What we know $(m)^{a/b}$=$(m^a)^{1/b}$ : always holds true, but we have different cases for negative $m$ that you will learn from my answer:
NOTE: $a/b$ is in simplest form.
Case 1: if $a$ is even, then $b$ will be odd which is sure, so u get a positive real value.
Case 2: if $a$ is odd and $b$ is also odd, so u get a negative real value.
Case 3: if $a$ is odd and $b$ is even, this is tricky here u get a undefined answer.
So for $Case 1$ and $Case 2$ $LHS=RHS$ always holds true and for third case you have undefined answer that you can read out why it is so.
See Images given below I have shown $Case 2$(that was asked by you) and $Case 3$ however you can try more on this.
Case 2:
Case 3:
kuspia
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We have that $(a)^{\frac{b}{c}}=(a^b)^{\frac{1}{c}}=(a^\frac1c)^{b}$ holds for any $a\ge 0$.
For the case $(-a)^{\frac{b}{c}}$ with $b$ and $c$ coprime we need to consider the following cases:
- $b$ odd and $c$ even
$$\left[(-a)^b\right]^\frac1c=\left[(-1)^b\cdot(a^b)\right]^\frac1c=\sqrt[c] {(-1)^b}\cdot a^{\frac{b}{c}} $$
$$\left[(-a)^\frac1c\right]^b=\left[(-1)^\frac1c \cdot a^\frac1c\right]^b=\left(\sqrt[c] {-1}\right)^b\cdot a^{\frac{b}{c}} $$
- $b$ even and $c$ odd
$$\left[(-a)^b\right]^\frac1c=\left[(-1)^b\cdot(a^b)\right]^\frac1c=\left[1\cdot(a^b)\right]^\frac1c=a^{\frac{b}{c}} $$
$$\left[(-a)^\frac1c\right]^b=\left[(-1)^\frac1c \cdot a^\frac1c\right]^b=\left(\sqrt[c] {-1}\right)^b\cdot(a)^{\frac{b}{c}} =(-1)^b \cdot (a)^{\frac{b}{c}}=a^{\frac{b}{c}}$$
- $b$ odd and $c$ odd
$$\left[(-a)^b\right]^\frac1c=\left[(-1)^b\cdot(a^b)\right]^\frac1c=\left[(-1)\cdot(a^b)\right]^\frac1c=\sqrt[c] {-1}\cdot(a)^{\frac{b}{c}}=-\left(a^{\frac{b}{c}}\right)$$
$$\left[(-a)^\frac1c\right]^b=\left[(-1)^\frac1c \cdot a^\frac1c\right]^b=\left[(-1) \cdot a^\frac1c\right]^b=(-1)^b\cdot(a)^{\frac{b}{c}} =-\left(a^{\frac{b}{c}}\right)$$
Therefore for the given example
$$(-2)^{\frac{7}{5}}=-\left(2^{\frac{7}{5}}\right)$$
user
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