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Popular >

solvefor x,h=2^x-1

Solution

solve for

x, h = 2^{x} - 1

Solution

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

Solution steps

h = 2^{x} - 1

Switch sides

2^{x} - 1 = h

Move

1

to the right side

2^{x} = h + 1

Apply exponent rules

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

Solve

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

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

Popular Examples

5^{x-1}+5^{x-2}=30

5^{x - 1} + 5^{x - 2} = 30

5^{x + 4} \cdot 25 = 125

130(5^{x-2})=1+25^x

130 (5^{x - 2}) = 1 + 2 5^{x}

(3/5)^x=7^{1-x}

(\frac{3}{5})^{x} = 7^{1 - x}

(10^2)/(10^x)=10

\frac{1 0 ^{2}}{1 0 ^{x}} = 10

Frequently Asked Questions (FAQ)

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