What is the answer to solvefor x,2^{x+1}-1=7^y ?

The answer to solvefor x,2^{x+1}-1=7^y is x=(ln(7^y+1))/(ln(2))-1

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solvefor x,2^{x+1}-1=7^y

solve for

x, 2^{x + 1} - 1 = 7^{y}

x = \frac{ln ( 7 ^{y} + 1 )}{ln ( 2 )} - 1

Solution steps

2^{x + 1} - 1 = 7^{y}

Move

1

to the right side

2^{x + 1} = 7^{y} + 1

Apply exponent rules

(x + 1) ln (2) = ln (7^{y} + 1)

Solve

(x + 1) ln (2) = ln (7^{y} + 1) : x = \frac{ln ( 7 ^{y} + 1 )}{ln ( 2 )} - 1

x = \frac{ln ( 7 ^{y} + 1 )}{ln ( 2 )} - 1

2^{4 x + 3} = 8^{x - 5}

5^{2 x + 1} = 3^{4 x - 1}

2 e^{0.5} x = e^{x}

a, x^{2} \cdot y^{3} = x^{a} \cdot y^{b}

2^{x + 2} = 5

What is the answer to solvefor x,2^{x+1}-1=7^y ?
The answer to solvefor x,2^{x+1}-1=7^y is x=(ln(7^y+1))/(ln(2))-1