Question

find x if (-1/3) raise to -18÷ (-1/3)raise to 7=(-1/3)raise to power -2x+1​

Answer 1

Question :-

Find x if $\rm \dfrac{( \frac{ - 1}{3} ) ^{ - 18}}{( \frac{ - 1}{3} ) ^{ 7}} = (\dfrac{ - 1}{3})^{ - 2x + 1}$.

Answer

➩ $\rm \dfrac{\Big( \frac{ - 1}{3}\Big) ^{ - 18}}{\Big( \frac{ - 1}{3} ) ^{ 7}} = (\dfrac{ - 1}{3}\Big)^{ - 2x + 1}$

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➩ $\rm \Big( \dfrac{ - 1}{3} \Big) ^{ - 18 - 7} = \Big(\frac{ - 1}{3} \Big) ^{ - 2x + 1}$

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➩ $\rm \Big(\dfrac{ - 1}{3} \Big)^{-25} = \Big(\dfrac{ - 1}{3}\Big)^{ - 2x + 1}$

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➩ $\rm -25 = -2x + 1$

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➩ $\rm 2x = 25 + 1$

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➩ $\rm 2x = 26$

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➩ $\rm x = \dfrac{26}{2}$

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➩ $\boxed{\rm x = 13 }$

Answer 2

Step-by-step explanation:

The answer is -1.

Lets's think about what is fractional exponent means. If the exponent is rational, we have

A^(B/C)

Which is the same as

(A^B)^(1/C)

Which is the same as

'C' root of (A^B)

If 'a' is negative, and 'b' is even, then a^b is positive, and we don't have a problem. On the other hand, if 'a' is negative, and 'b' is odd, the a^b is negative, and that's a problem is 'c' is even.

When i say 'Problem', I means a problem if we're limited to working with real numbers. But if we extend our number system to include complex numbers, then we have

(-1)^(1/2) = Square root((-1)^1) = Sqrt(-1) = i