What is the answer to log_{3}(x)=1-x/2 ?

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

Solve for x: log_{3}(x)=1-x/2

The solution is x=(2\W_{0}((3ln(3))/2))/(ln(3))

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

x = \frac{2 W _{0} ( \frac{3 l n ( 3 )}{2} )}{ln ( 3 )}

Decimal Notation

x = 1.39455 \dots

Solution steps

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

f (x) = g (x),

then

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

3^{l o g_{3} (x)} = 3^{1 - \frac{x}{2}}

Simplify

3^{l o g_{3} (x)} : x

x = 3^{1 - \frac{x}{2}}

Solve

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

x = \frac{2 W _{0} ( \frac{3 l n ( 3 )}{2} )}{ln ( 3 )}

Verify Solutions:

x = \frac{2 W _{0} ( \frac{3 l n ( 3 )}{2} )}{ln ( 3 )}

True

The solution is

x = \frac{2 W _{0} ( \frac{3 l n ( 3 )}{2} )}{ln ( 3 )}

5 lo g_{5} (x) + 5 = 0

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

lo g_{32} (x) = - \frac{2}{5}

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

lo g_{5} (2 x - 5) = 0

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