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Popular >

solvefor x,y=2^x-1

Solution

solve for

x, y = 2^{x} - 1

Solution

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

Solution steps

y = 2^{x} - 1

Switch sides

2^{x} - 1 = y

Move

1

to the right side

2^{x} = y + 1

Apply exponent rules

x ln (2) = ln (y + 1)

Solve

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

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

Popular Examples

64 = 2 e^{8 x} - 8

4*2^{2x}+3*2^x-1=0

4 \cdot 2^{2 x} + 3 \cdot 2^{x} - 1 = 0

0 = 2. 3^{x} + 1

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

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

5^{x^{2}} = 5

Frequently Asked Questions (FAQ)

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