What is the answer to log_{2}(x-1)=-2022 ?

The answer to log_{2}(x-1)=-2022 is x= 1/(2^{2022)}+1

Solve for x: log_{2}(x-1)=-2022

The solution is x= 1/(2^{2022)}+1

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lo g_{2} (x - 1) = - 2022

x = \frac{1}{2 ^{2022}} + 1

Solution steps

lo g_{2} (x - 1) = - 2022

Apply log rules

x - 1 = \frac{1}{2 ^{2022}}

Solve

x - 1 = \frac{1}{2 ^{2022}} : x = \frac{1}{2 ^{2022}} + 1

x = \frac{1}{2 ^{2022}} + 1

(lo g_{2} (x))^{2} - 5 \cdot lo g_{2} (x) + 6 = 0

lo g_{10} (a) = 6

2 lo g_{10} (3 x) + lo g_{10} (18 x) = 27

lo g_{6} (x) + lo g_{6} (x + 5) = 2

lo g_{10} (x) + lo g_{10} (2 x - 1) = lo g_{10} (6)

What is the answer to log_{2}(x-1)=-2022 ?
The answer to log_{2}(x-1)=-2022 is x= 1/(2^{2022)}+1
Solve for x: log_{2}(x-1)=-2022
The solution is x= 1/(2^{2022)}+1