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.
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$\begingroup$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.
$\endgroup$ $\begingroup$$$5\cdot125^{\frac{1}{x}}=\left(\frac{1}{5}\right)^{\frac{1}{x}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1^{\frac{1}{x}}}{5^{\frac{1}{x}}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow$$
Use (real solution) $5^y=125\Longleftrightarrow y=\log_5(125)=3$:
$$5\cdot(5^3)^{\frac{1}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5\cdot5^{\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5^1\cdot5^{\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow$$ $$5^{1+\frac{3}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow5^{1+\frac{3}{x}}\cdot5^{\frac{1}{x}}=1\Longleftrightarrow5^{1+\frac{1}{x}+\frac{3}{x}}=1\Longleftrightarrow$$ $$5^{1+\frac{4}{x}}=1\Longleftrightarrow\log_5\left(5^{1+\frac{4}{x}}\right)=\log_5(1)\Longleftrightarrow1+\frac{4}{x}=0\Longleftrightarrow x=-4$$
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