What is the answer to log_{2}(x-1)+x=1 ?

The answer to log_{2}(x-1)+x=1 is x=(\W_{0}(ln(2)))/(ln(2))+1

Solve for x: log_{2}(x-1)+x=1

The solution is x=(\W_{0}(ln(2)))/(ln(2))+1

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

x = \frac{W _{0} ( ln ( 2 ) )}{ln ( 2 )} + 1

Decimal Notation

x = 1.64118 \dots

Solution steps

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

Move

x

to the right side

lo g_{2} (x - 1) = 1 - x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

2^{l o g_{2} (x - 1)} = 2^{1 - x}

Simplify

2^{l o g_{2} (x - 1)} : x - 1

x - 1 = 2^{1 - x}

Solve

x - 1 = 2^{1 - x} : x = \frac{W _{0} ( ln ( 2 ) )}{ln ( 2 )} + 1

x = \frac{W _{0} ( ln ( 2 ) )}{ln ( 2 )} + 1

Verify Solutions:

x = \frac{W _{0} ( ln ( 2 ) )}{ln ( 2 )} + 1

True

The solution is

x = \frac{W _{0} ( ln ( 2 ) )}{ln ( 2 )} + 1

lo g_{x} (x^{2}) = lo g_{8} (16 x)

lo g_{10} (x) = 18

lo g_{10} (x) (8 - 2 x) = 2

lo g_{10} (x) + x = 0

lo g_{x} (5) - lo g_{25} (x) - 1 = 0

What is the answer to log_{2}(x-1)+x=1 ?
The answer to log_{2}(x-1)+x=1 is x=(\W_{0}(ln(2)))/(ln(2))+1
Solve for x: log_{2}(x-1)+x=1
The solution is x=(\W_{0}(ln(2)))/(ln(2))+1