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

The answer to 2log_{10}(x+1)=2-x is x=(2\W_{0}((ln(10)e^{\frac{3ln(10))/2})/(2))}{ln(10)}-1

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

The solution is x=(2\W_{0}((ln(10)e^{\frac{3ln(10))/2})/(2))}{ln(10)}-1

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

x = \frac{2 W _{0} ( \frac{l n ( 10 ) e ^{\frac{3 l n ( 10 )}{2}}}{2} )}{ln ( 10 )} - 1

Solution steps

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

Divide both sides by

2

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

f (x) = g (x),

then

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

1 0^{l o g_{10} (x + 1)} = 1 0^{\frac{2 - x}{2}}

Simplify

1 0^{l o g_{10} (x + 1)} : x + 1

x + 1 = 1 0^{\frac{2 - x}{2}}

Solve

x + 1 = 1 0^{\frac{2 - x}{2}} : x = \frac{2 W _{0} ( \frac{l n ( 10 ) e ^{\frac{3 l n ( 10 )}{2}}}{2} )}{ln ( 10 )} - 1

x = \frac{2 W _{0} ( \frac{l n ( 10 ) e ^{\frac{3 l n ( 10 )}{2}}}{2} )}{ln ( 10 )} - 1

Verify Solutions:

x = \frac{2 W _{0} ( \frac{l n ( 10 ) e ^{\frac{3 l n ( 10 )}{2}}}{2} )}{ln ( 10 )} - 1

True

The solution is

x = \frac{2 W _{0} ( \frac{l n ( 10 ) e ^{\frac{3 l n ( 10 )}{2}}}{2} )}{ln ( 10 )} - 1

lo g_{6} (x) + lo g_{6} (x + 1) = 1

lo g_{10} (a + 4) = 5.2

lo g_{10} (10 x) = 4

ln (x^{2} - 1) - ln (x - 1) = 2

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

What is the answer to 2log_{10}(x+1)=2-x ?
The answer to 2log_{10}(x+1)=2-x is x=(2\W_{0}((ln(10)e^{\frac{3ln(10))/2})/(2))}{ln(10)}-1
Solve for x: 2log_{10}(x+1)=2-x
The solution is x=(2\W_{0}((ln(10)e^{\frac{3ln(10))/2})/(2))}{ln(10)}-1