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For the equation

$$5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $$

$x$ is equal to $-4$, but I'm not sure why.

I've taken the right side of the equation ${ \left( \frac { 1 }{ 5 } \right) }^{ \frac { 1 }{ x } }$ and converted it to $5^{-(1/x)}$, but haven't been able to perform the mathematical gymnastics to prove this.

Zain Patel
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maudulus
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    The negative power rule states $1/a^b=a^{-b}$, which you applied correctly. Now you need to rewrite the left side as a single power of $5$ in order to employ the method of relating bases. You should get $$5^{1+3/x}=5^{-1/x}.$$ From there, set the exponents equal to each other ($1+3/x=-1/x$) and solve. – anon Jul 21 '16 at 18:02
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    I suggest looking at MathJax to format your question better – Joshua Lochner Jul 21 '16 at 18:05
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    I don't think it's a good idea to edit OP's original equations using exponent rules when this is the level of math where they're first learning exponent rules. I have edited it back to original form (but with $\LaTeX$). – anon Jul 21 '16 at 18:07
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    @Battani why did you edit his original equation like that? Leave it how the question was given – Joshua Lochner Jul 21 '16 at 18:09
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    actually, the formatting that Battani did was good, I just didn't know how to write MathJax – maudulus Jul 21 '16 at 18:11
  • @maudulus Did your original problem write $\sqrt[x]{125}$ or did it write $125^{1/x}$? – anon Jul 21 '16 at 18:12
  • @JoshuaLochner,it is not polite when you order,what was wrong? – haqnatural Jul 21 '16 at 18:13

2 Answers2

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Since $125 = 5^3$, we have $$5 \cdot 5^{\frac{3}{x}} = \frac{1}{5^{\frac{1}{x}}} \Rightarrow 5^{1 + \frac{3}{x}} = 5^{-\frac{1}{x}}$$

Equating powers gives $1 + \frac{3}{x} = -\frac{1}{x} \Rightarrow x = -4$.

Some explanations:

We have $a^b \cdot a^c = a^{b+c}$, this can be reasoned as such: $$a^b \cdot a^c = \underbrace{a \cdot a \cdots a}_{b \, \text{times}} \cdot \underbrace{a \cdot a \cdots a}_{c \, \text{times}} = \underbrace{a \cdot a \cdots a}_{(b+c) \, \text{times}} = a^{b+c}$$ This gives $5 \cdot 5^{3/x} = 5^{1 + 3/x}$ as promised.

Next, we have $(a^b)^c = (a^c)^b = a^{bc}$, this should make sense if you think of it being $a^b$ multiplied together $c$ times, giving a total of $a$ being multiplied together $bc$ times, since $a^b$ is $a$ multiplied together $b$ times. Symbolically $$(a^b)^c = \underbrace{a^b \cdot a^b \cdots a^b}_{c \, \text{times}} = a^{\overbrace{b + \cdots + b}^{c \, \text{times}}} = a^{bc}$$

This is what lets us write $125^{1/x} = (5^3)^{1/x} = 5^{3/x}$.

I do need to emphasise that my 'explanations' are simply that, visual and handy intuitive explanations, they do not constitute proper proof (as we are implicitly assuming $a,b,c$ are (positive) integers, whilst the stated rules hold for real numbers.

Zain Patel
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0

$$5\cdot125^{\frac{1}{x}}=\left(\frac{1}{5}\right)^{\frac{1}{x}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1^{\frac{1}{x}}}{5^{\frac{1}{x}}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow$$

Use (real solution) $5^y=125\Longleftrightarrow y=\log_5(125)=3$:

$$5\cdot(5^3)^{\frac{1}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5\cdot5^{\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5^1\cdot5^{\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow$$ $$5^{1+\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5^{1+\frac{3}{x}}\cdot5^{\frac{1}{x}}=1\Longleftrightarrow5^{1+\frac{1}{x}+\frac{3}{x}}=1\Longleftrightarrow$$ $$5^{1+\frac{4}{x}}=1\Longleftrightarrow\log_5\left(5^{1+\frac{4}{x}}\right)=\log_5(1)\Longleftrightarrow1+\frac{4}{x}=0\Longleftrightarrow x=-4$$

Jan Eerland
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