What is the answer to 40=-10log_{10}((1-x)^2) ?

The answer to 40=-10log_{10}((1-x)^2) is x= 99/100 ,x= 101/100

Solve for x: 40=-10log_{10}((1-x)^2)

The solution is x= 99/100 ,x= 101/100

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40 = - 10 lo g_{10} ((1 - x)^{2})

x = \frac{99}{100}, x = \frac{101}{100}

Solution steps

40 = - 10 lo g_{10} ((1 - x)^{2})

Switch sides

- 10 lo g_{10} ((1 - x)^{2}) = 40

Divide both sides by

- 10

lo g_{10} ((1 - x)^{2}) = - 4

Apply log rules

(1 - x)^{2} = \frac{1}{10000}

Solve

(1 - x)^{2} = \frac{1}{10000} : x = \frac{99}{100}, x = \frac{101}{100}

x = \frac{99}{100}, x = \frac{101}{100}

Verify Solutions:

x = \frac{99}{100}

True

, x = \frac{101}{100}

True

The solutions are

x = \frac{99}{100}, x = \frac{101}{100}

lo g_{2} (x^{3} - 56) = 3

ln (x - 1) + ln (2) = 0

lo g_{2} (x + 2) + lo g_{2} (4) = 4

lo g_{16} (x) = - 4^{- 1}

ln (x - 2) = 6

What is the answer to 40=-10log_{10}((1-x)^2) ?
The answer to 40=-10log_{10}((1-x)^2) is x= 99/100 ,x= 101/100
Solve for x: 40=-10log_{10}((1-x)^2)
The solution is x= 99/100 ,x= 101/100