What is the answer to 3= 5/2 x+log_{e}(2x) ?

The answer to 3= 5/2 x+log_{e}(2x) is x=(2\W_{0}((5e^3)/4))/(5)

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3= 5/2 x+log_{e}(2x)

3 = \frac{5}{2} x + lo g_{e} (2 x)

x = \frac{2 W _{0} ( \frac{5 e ^{3}}{4} )}{5}

Decimal Notation

x = 0.94525 \dots

Solution steps

3 = \frac{5}{2} x + lo g_{e} (2 x)

Switch sides

\frac{5}{2} x + lo g_{e} (2 x) = 3

Move

\frac{5}{2} x

to the right side

lo g_{e} (2 x) = 3 - \frac{5}{2} x

f (x) = g (x),

then

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

e^{l o g_{e} (2 x)} = e^{3 - \frac{5}{2} x}

Simplify

e^{l o g_{e} (2 x)} : 2 x

2 x = e^{3 - \frac{5}{2} x}

Solve

2 x = e^{3 - \frac{5}{2} x} : x = \frac{2 W _{0} ( \frac{5 e ^{3}}{4} )}{5}

x = \frac{2 W _{0} ( \frac{5 e ^{3}}{4} )}{5}

Verify Solutions:

x = \frac{2 W _{0} ( \frac{5 e ^{3}}{4} )}{5}

True

The solution is

x = \frac{2 W _{0} ( \frac{5 e ^{3}}{4} )}{5}

lo g_{4} (x^{2} + 2) = 3

32 = 20 lo g_{10} (x)

ln (x) = 3.68193

2 lo g_{3} (7 x - 11) + 6 = 10

lo g_{10} (x^{2} - 3 x + 6) = 1

What is the answer to 3= 5/2 x+log_{e}(2x) ?
The answer to 3= 5/2 x+log_{e}(2x) is x=(2\W_{0}((5e^3)/4))/(5)