What is the answer to solvefor x,4^x-2(8^y)=0 ?

The answer to solvefor x,4^x-2(8^y)=0 is x=(3y+1)/2

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solvefor x,4^x-2(8^y)=0

solve for

x, 4^{x} - 2 (8^{y}) = 0

x = \frac{3 y + 1}{2}

Solution steps

4^{x} - 2 (8^{y}) = 0

Add

2 \cdot 8^{y}

to both sides

4^{x} - 2 \cdot 8^{y} + 2 \cdot 8^{y} = 0 + 2 \cdot 8^{y}

Simplify

0 + 2 \cdot 8^{y} : 2^{3 y + 1}

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

Apply exponent rules

2 x = 3 y + 1

Solve

2 x = 3 y + 1 : x = \frac{3 y + 1}{2}

x = \frac{3 y + 1}{2}

3^{m + 2} - 2 \cdot 3^{m} = 63

1 1^{x - 3} = 133 1^{x - 3}

2^{x^{3}} = 2^{3^{6}}

m, m 2^{m} + n 2^{n} = 32768

3^{2 x - 2} - 5 \cdot 3^{x - 2} - 66 = 0

What is the answer to solvefor x,4^x-2(8^y)=0 ?
The answer to solvefor x,4^x-2(8^y)=0 is x=(3y+1)/2