What is the answer to 2^{n-1}+2^{n-2}+2^{n-3}=896 ?

The answer to 2^{n-1}+2^{n-2}+2^{n-3}=896 is n=10

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2^{n-1}+2^{n-2}+2^{n-3}=896

2^{n - 1} + 2^{n - 2} + 2^{n - 3} = 896

n = 10

Solution steps

2^{n - 1} + 2^{n - 2} + 2^{n - 3} = 896

Factor

2^{n - 1} + 2^{n - 2} + 2^{n - 3} : 7 \cdot 2^{- 3 + n}

7 \cdot 2^{- 3 + n} = 896

Divide both sides by

7

2^{- 3 + n} = 128

Apply exponent rules

- 3 + n = 7

Solve

- 3 + n = 7 : n = 10

n = 10

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

5^{x} = \frac{1}{25}

\frac{1}{( 3 ^{- 2 x + 2} )} = 81

1 6^{p} + 4^{p - 1} = \frac{1}{4 ^{3}}

74 = 1 0^{2.37 x}

What is the answer to 2^{n-1}+2^{n-2}+2^{n-3}=896 ?
The answer to 2^{n-1}+2^{n-2}+2^{n-3}=896 is n=10
Solve for n: 2^{n-1}+2^{n-2}+2^{n-3}=896
The solution is n=10