What is the answer to solvefor x,f=1+log_{2}(x+k) ?

The answer to solvefor x,f=1+log_{2}(x+k) is x=2^{f-1}-k

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solvefor x,f=1+log_{2}(x+k)

solve for

x, f = 1 + lo g_{2} (x + k)

x = 2^{f - 1} - k

Solution steps

f = 1 + lo g_{2} (x + k)

Switch sides

1 + lo g_{2} (x + k) = f

Move

1

to the right side

lo g_{2} (x + k) = f - 1

Apply log rules

x + k = 2^{f - 1}

Solve

x + k = 2^{f - 1} : x = 2^{f - 1} - k

x = 2^{f - 1} - k

lo g_{2} (N) = - 2

0 = ln (x) + 1

0 = ln (x) + 2

2 ln (6 x - 3) + 11 = 19

lo g_{2} (2) + lo g_{2} (x) = 2

What is the answer to solvefor x,f=1+log_{2}(x+k) ?
The answer to solvefor x,f=1+log_{2}(x+k) is x=2^{f-1}-k