What is the answer to 2+log_{10}(1-x)=0 ?

The answer to 2+log_{10}(1-x)=0 is x= 99/100

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2+log_{10}(1-x)=0

2 + lo g_{10} (1 - x) = 0

x = \frac{99}{100}

Solution steps

2 + lo g_{10} (1 - x) = 0

Move

2

to the right side

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

Apply log rules

1 - x = \frac{1}{100}

Solve

1 - x = \frac{1}{100} : x = \frac{99}{100}

x = \frac{99}{100}

Verify Solutions:

x = \frac{99}{100}

True

The solution is

x = \frac{99}{100}

3 + lo g_{2} (2 x) = lo g_{2} (17 x - 5)

ln (x - 4) - ln (x) = 5

x, ln (y) = 2 ln (x) + 2 c

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

lo g_{e} (x - 2) - 3 lo g_{e} (2) = 2

What is the answer to 2+log_{10}(1-x)=0 ?
The answer to 2+log_{10}(1-x)=0 is x= 99/100
Solve for x: 2+log_{10}(1-x)=0
The solution is x= 99/100