What is the answer to 2^k-1+2^{(k+1)-1}=2^{k+1}-1 ?

The answer to 2^k-1+2^{(k+1)-1}=2^{k+1}-1 is True for all k

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2^k-1+2^{(k+1)-1}=2^{k+1}-1

2^{k} - 1 + 2^{(k + 1) - 1} = 2^{k + 1} - 1

True for all k

Solution steps

2^{k} - 1 + 2^{(k + 1) - 1} = 2^{k + 1} - 1

Subtract

2^{k + 1}

from both sides

2^{k} - 1 + 2^{k + 1 - 1} - 2^{k + 1} = 2^{k + 1} - 1 - 2^{k + 1}

Simplify

2^{k} - 1 + 2^{k + 1 - 1} - 2^{k + 1} = - 1

Factor

2^{k} - 1 + 2^{k + 1 - 1} - 2^{k + 1} : - 1

- 1 = - 1

Both sides are equal

True for all k

L, T = e^{- k L}

9^{x} + 2 \cdot 3^{x + 1} = 27

2 5^{z + 4} = 125

3^{4 x} = 7^{x^{2}}

5^{x^{2}} = 2 5^{3 x - 4}

What is the answer to 2^k-1+2^{(k+1)-1}=2^{k+1}-1 ?
The answer to 2^k-1+2^{(k+1)-1}=2^{k+1}-1 is True for all k
Solve for k: 2^k-1+2^{(k+1)-1}=2^{k+1}-1
The solution is True for all k