What is the answer to log_{10}(x)-4((1-x)/2)=1 ?

The answer to log_{10}(x)-4((1-x)/2)=1 is x=(\W_{0}(2000ln(10)))/(2ln(10))

Solve for x: log_{10}(x)-4((1-x)/2)=1

The solution is x=(\W_{0}(2000ln(10)))/(2ln(10))

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

x = \frac{W _{0} ( 2000 ln ( 10 ) )}{2 ln ( 10 )}

Decimal Notation

x = 1.42334 \dots

Solution steps

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

4 \cdot \frac{1 - x}{2} = 2 (- x + 1)

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

Move

2 (- x + 1)

to the right side

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

f (x) = g (x),

then

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

1 0^{l o g_{10} (x)} = 1 0^{- 2 x + 3}

Simplify

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

x = 1 0^{- 2 x + 3}

Solve

x = 1 0^{- 2 x + 3} : x = \frac{W _{0} ( 2000 ln ( 10 ) )}{2 ln ( 10 )}

x = \frac{W _{0} ( 2000 ln ( 10 ) )}{2 ln ( 10 )}

Verify Solutions:

x = \frac{W _{0} ( 2000 ln ( 10 ) )}{2 ln ( 10 )}

True

The solution is

x = \frac{W _{0} ( 2000 ln ( 10 ) )}{2 ln ( 10 )}

ln (2 x + 1) - ln (x - 6) = 0

lo g_{10} (x) = 5

lo g_{2} (2 x + 7) = 4

lo g_{5} (x) = 1 + lo g_{10} (x^{25})

\frac{x}{lo g _{2} ( 4 x - 12 )} = 5

What is the answer to log_{10}(x)-4((1-x)/2)=1 ?
The answer to log_{10}(x)-4((1-x)/2)=1 is x=(\W_{0}(2000ln(10)))/(2ln(10))
Solve for x: log_{10}(x)-4((1-x)/2)=1
The solution is x=(\W_{0}(2000ln(10)))/(2ln(10))