What is the answer to 4^{sqrt(x+1)}=2^{3x-2} ?

The answer to 4^{sqrt(x+1)}=2^{3x-2} is x= 16/9

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4^{sqrt(x+1)}=2^{3x-2}

4^{x + 1} = 2^{3 x - 2}

x = \frac{16}{9}

Decimal Notation

x = 1.77777 \dots

Solution steps

4^{x + 1} = 2^{3 x - 2}

Apply exponent rules

2 x + 1 = 3 x - 2

Solve

2 x + 1 = 3 x - 2 : x = \frac{16}{9}

x = \frac{16}{9}

3^{x} = 7

4 e^{2 x} - 7 e^{x} = 15

2^{2 x} - 5 \cdot 2^{x} + 6 = 0

7^{x + 1} - 7^{x - 1} = 48

e^{e^{x}} = 2

What is the answer to 4^{sqrt(x+1)}=2^{3x-2} ?
The answer to 4^{sqrt(x+1)}=2^{3x-2} is x= 16/9
Solve for x: 4^{sqrt(x+1)}=2^{3x-2}
The solution is x= 16/9