What is the answer to 4^n+4^n+4^n+4^n=2^{44} ?

The answer to 4^n+4^n+4^n+4^n=2^{44} is n=21

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4^n+4^n+4^n+4^n=2^{44}

4^{n} + 4^{n} + 4^{n} + 4^{n} = 2^{44}

n = 21

Solution steps

4^{n} + 4^{n} + 4^{n} + 4^{n} = 2^{44}

Add similar elements:

4^{n} + 4^{n} + 4^{n} + 4^{n} = 4 \cdot 4^{n}

4 \cdot 4^{n} = 2^{44}

Divide both sides by

4

4^{n} = \frac{2 ^{44}}{4}

Simplify

\frac{2 ^{44}}{4} : 2^{42}

4^{n} = 2^{42}

Apply exponent rules

2 n = 42

Solve

2 n = 42 : n = 21

n = 21

9^{x} + 9^{x} = 9

4^{x - 3} = 0. 5^{x}

(\frac{1}{4})^{4 x} = 64

3^{5 x - 4} = 2 7^{x + 2}

4^{2 x + 1} = 17 (4^{x}) - 15

What is the answer to 4^n+4^n+4^n+4^n=2^{44} ?
The answer to 4^n+4^n+4^n+4^n=2^{44} is n=21
Solve for n: 4^n+4^n+4^n+4^n=2^{44}
The solution is n=21