Say I have the following expression:
$10^{l} = a^{k}$
If I take the kth root of both sides, does that mean we get:
$10^{\frac{l}{k}} = a$
We don't have to consider anything with plus or minus?
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2$\begingroup$ Suppose $10^8=a^4.$ Try plugging in $a=100.$ Try $a=-100.$ What do you think? $\endgroup$– David KCommented Jun 18 at 0:25
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$\begingroup$ So only when k is even do we consider plus and minuses? $\endgroup$– Bob MarleyCommented Jun 18 at 0:30
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1$\begingroup$ What do the graphs of $y=x$, $y=x^2$, $y=x^3$, $y=x^4$, etc. look like? $\endgroup$– Paul TanenbaumCommented Jun 18 at 3:40
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$\begingroup$ It is tricky. There are many points to consider. For example, are the powers real? are the bases real? are the bases all positive? This may be helpful: math.stackexchange.com/questions/4640732/… $\endgroup$– NoChanceCommented Jun 18 at 11:19
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1 Answer
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Say $a^k = b$ where $a, b$ are real numbers and $k$ is an integer.
This really just depends on whether $k$ is even or odd.
Say $k$ is even. Then $a^k = b > 0$ regardless of $a$ positive or negative. But in the same way $b > 0$ means that $b^{\frac{1}{k}}$ can be either positive or negative since putting this root to an even power will negate the sign.
If $k$ is odd however, then $a^k = b > 0$ only if $a > 0$ and $a^k = b < 0$ only if $a < 0$, i.e. the sign of $b$ matches the sign of $a$. And because $k$ is odd, $b^{\frac{1}{k}}$ must have the same sign as $b$ and a priori as $a$ also.
So now in your example, you know that $10^l > 0$ regardless of $l$. So...
