What is the answer to (2^8}{3^8}=(\frac{-2)/3)^{n-1} ?

The answer to (2^8}{3^8}=(\frac{-2)/3)^{n-1} is n=9

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(2^8}{3^8}=(\frac{-2)/3)^{n-1}

\frac{2 ^{8}}{3 ^{8}} = (\frac{- 2}{3})^{n - 1}

n = 9

Solution steps

\frac{2 ^{8}}{3 ^{8}} = (\frac{- 2}{3})^{n - 1}

Apply exponent rules

2^{8} \cdot 3^{- 8} = (\frac{- 2}{3})^{n - 1}

(- a)^{n} = b,

assuming

n

is even, solve

a^{n} = b

instead

Solve

(\frac{2}{3})^{n - 1} = 2^{8} \cdot 3^{- 8} : n = 9

Verify Solutions:

n = 9

True

The solution is

n = 9

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4^{x + 2} + 4^{x} = 1088

2^{x} = 100

1 5^{- x + 8} = 1 4^{7 x}

What is the answer to (2^8}{3^8}=(\frac{-2)/3)^{n-1} ?
The answer to (2^8}{3^8}=(\frac{-2)/3)^{n-1} is n=9
Solve for n: (2^8}{3^8}=(\frac{-2)/3)^{n-1}
The solution is n=9