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How would you solve for $3^{-4}$? I know the answer is $1/81$ but I can't work out how you get there with this one.

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  • $\begingroup$ Can you do 3^4? and after that a^-1 = 1/a ? $\endgroup$ – novice May 31 '16 at 10:05
  • $\begingroup$ $3^{-4}\cdot 3^4= 3^0=1$ $\endgroup$ – Piquito May 31 '16 at 14:55
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$3^{-4}=({3^4})^{-1}$

But $3^4=(3^2)^2=9^2=81$

So we get the answer is $81^{-1}=\frac{1}{81}$

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Well we know $3^4$ equals 81, because $3 \times 3 \times 3 \times 3 = 81$

and there is a function where any value to the power of a minus become the inverse of that number.

An inverse is basically that number turned upside down.

For example: $x^{-1}$ become $1/x$ and $2^{-1} $ becomes 1/2

Therefore with these two things in mind we have $81^{-1}$

Giving your answer :^) 1/81

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$3^4$ is the abbreviation for multiplying four times by $3$, that is multiply by $81$. Now $3^{-4}$ is the abbreviation for dividing four times by $3$, that is, multiply by $1/81$.

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Let's consider the expression $$\frac{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3}$$ There are three more $3$s multiplied together on the top than there are on the bottom, and we write it more concisely as $3^3$.

Suppose now that the number of $3$s on top are equal to the number of threes on the bottom. In this case we would write $3^0$ (which simplifies to just $1$). $$\frac{3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3}$$

Following this model, we would expect that $3$ raised to a negative power would represent a fraction where the number of $3$s on the bottom exceed the number of $3$s on the top, as follows. $$\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}$$which we would represent as $3^{-4}$.

Continuing we have$$\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3}=\frac{3\cdot 3}{3\cdot 3\cdot ((3\cdot 3)\cdot (3\cdot 3))}=\frac{3\cdot 3}{3\cdot 3\cdot (9\cdot 9)}=\frac{1}{81}$$

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Well $3^{-4}$ = $3^{0-4}$

therefore, $3^{0-4}$ = $3^0/3^4$ [As we know $x^{a-b}$ = $x^a/x^b$]

therefore, $3^{0-4}$ = $3^0/3^4$ = $1/81$

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  • $\begingroup$ That "As we know" statement is for positive integral exponents. $\endgroup$ – pushpen.paul Jun 8 '16 at 8:15
  • $\begingroup$ Didn't mention it as the context is on positive integral exponents $\endgroup$ – Hailey Jun 8 '16 at 8:17
  • $\begingroup$ Doesn't that mean you are giving a new definition? You can't 'prove' this state except when both exponents are positive integer and the subtraction result is positive. When you use $3^{0-4}=3^0/3^4$, you are assuming/generalizing the definition. $\endgroup$ – pushpen.paul Jun 8 '16 at 15:24
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you can think like this let $3^{-4}=a$ where a could be any real number. and we know a property $m^{-n}=1/m^n$ therefore$$3^{-4}=a\implies\frac 1{3^4}=a $$$$1=a\times3^4$$now think of a value of $a$ which when multiplied by $3^4$ gives 1. yes it is $1/3^4$

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