What is the answer to ln(5x)=2x-1 ?

The answer to ln(5x)=2x-1 is x=-(\W_{-1}(-2/(5e)))/2 ,x=-(\W_{0}(-2/(5e)))/2

Solve for x: ln(5x)=2x-1

The solution is x=-(\W_{-1}(-2/(5e)))/2 ,x=-(\W_{0}(-2/(5e)))/2

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ln (5 x) = 2 x - 1

x = - \frac{W _{- 1} ( - \frac{2}{5 e} )}{2}, x = - \frac{W _{0} ( - \frac{2}{5 e} )}{2}

Decimal

x = 1.51115 \dots, x = 0.08767 \dots

Solution steps

ln (5 x) = 2 x - 1

f (x) = g (x),

then

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

e^{l n (5 x)} = e^{2 x - 1}

Simplify

e^{l n (5 x)} : 5 x

5 x = e^{2 x - 1}

Solve

5 x = e^{2 x - 1} : x = - \frac{W _{- 1} ( - \frac{2}{5 e} )}{2}, x = - \frac{W _{0} ( - \frac{2}{5 e} )}{2}

x = - \frac{W _{- 1} ( - \frac{2}{5 e} )}{2}, x = - \frac{W _{0} ( - \frac{2}{5 e} )}{2}

Verify Solutions:

x = - \frac{W _{- 1} ( - \frac{2}{5 e} )}{2}

True

, x = - \frac{W _{0} ( - \frac{2}{5 e} )}{2}

True

The solutions are

x = - \frac{W _{- 1} ( - \frac{2}{5 e} )}{2}, x = - \frac{W _{0} ( - \frac{2}{5 e} )}{2}

ln (x) = 7

ln ((x + 1)^{2}) = 2

(x - 1) ln (x) = 0

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

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

What is the answer to ln(5x)=2x-1 ?
The answer to ln(5x)=2x-1 is x=-(\W_{-1}(-2/(5e)))/2 ,x=-(\W_{0}(-2/(5e)))/2
Solve for x: ln(5x)=2x-1
The solution is x=-(\W_{-1}(-2/(5e)))/2 ,x=-(\W_{0}(-2/(5e)))/2