What is the answer to 2log_{10}(9-2x)=(3-x) ?

The answer to 2log_{10}(9-2x)=(3-x) is

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2log_{10}(9-2x)=(3-x)

2 lo g_{10} (9 - 2 x) = (3 - x)

x = \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}, x = \frac{2 W _{0} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}

Decimal Notation

x = 1.42116 \dots, x = 4.40026 \dots

Solution steps

2 lo g_{10} (9 - 2 x) = (3 - x)

Divide both sides by

2

lo g_{10} (9 - 2 x) = \frac{3 - x}{2}

f (x) = g (x),

then

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

1 0^{l o g_{10} (9 - 2 x)} = 1 0^{\frac{3 - x}{2}}

Simplify

1 0^{l o g_{10} (9 - 2 x)} : 9 - 2 x

9 - 2 x = 1 0^{\frac{3 - x}{2}}

Solve

9 - 2 x = 1 0^{\frac{3 - x}{2}} : x = \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}, x = \frac{2 W _{0} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}

x = \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}, x = \frac{2 W _{0} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}

Verify Solutions:

x = \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}

True

, x = \frac{2 W _{0} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}

True

The solutions are

x = \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}, x = \frac{2 W _{0} ( - \frac{l n ( 10 )}{4 e ^{\frac{3 l n ( 10 )}{4}}} )}{ln ( 10 )} + \frac{9}{2}

lo g_{10} (5 x + 1) = 2 + lo g_{10} (2 x - 3)

4 lo g_{2} (2 x) = 36

lo g_{10} (n) = - 0.35

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

lo g_{2} (x) = lo g_{4} (x^{2} + 2 x - 1)

What is the answer to 2log_{10}(9-2x)=(3-x) ?
The answer to 2log_{10}(9-2x)=(3-x) is
Solve for x: 2log_{10}(9-2x)=(3-x)
The solution is