What is the answer to ln(x^2)+2x-5=0 ?

The answer to ln(x^2)+2x-5=0 is x=\W_{0}(e^{5/2})

Solve for x: ln(x^2)+2x-5=0

The solution is x=\W_{0}(e^{5/2})

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ln (x^{2}) + 2 x - 5 = 0

x = W_{0} (e^{\frac{5}{2}})

Decimal Notation

x = 1.87264 \dots

Solution steps

ln (x^{2}) + 2 x - 5 = 0

Move

2 x

to the right side

ln (x^{2}) - 5 = - 2 x

Move

5

to the right side

ln (x^{2}) = - 2 x + 5

f (x) = g (x),

then

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

e^{l n (x^{2})} = e^{- 2 x + 5}

Simplify

e^{l n (x^{2})} : x^{2}

x^{2} = e^{- 2 x + 5}

Solve

x^{2} = e^{- 2 x + 5} : x = W_{0} (e^{\frac{5}{2}})

x = W_{0} (e^{\frac{5}{2}})

Verify Solutions:

x = W_{0} (e^{\frac{5}{2}})

True

The solution is

x = W_{0} (e^{\frac{5}{2}})

ln (x) = 4 lo g_{x} (e)

lo g_{9} (4) = 2 lo g_{9} (x)

lo g_{2} (x) = lo g_{2} (3 x - 1)

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

5 ln (8 x - 5) - 20 = 5

What is the answer to ln(x^2)+2x-5=0 ?
The answer to ln(x^2)+2x-5=0 is x=\W_{0}(e^{5/2})
Solve for x: ln(x^2)+2x-5=0
The solution is x=\W_{0}(e^{5/2})