What is the answer to ln(2x)+ln(4x)=0 ?

The answer to ln(2x)+ln(4x)=0 is x= 1/(e^{\frac{3ln(2)){2}}}

Solve for x: ln(2x)+ln(4x)=0

The solution is x= 1/(e^{\frac{3ln(2)){2}}}

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ln (2 x) + ln (4 x) = 0

x = \frac{1}{e ^{\frac{3 l n ( 2 )}{2}}}

Solution steps

ln (2 x) + ln (4 x) = 0

Apply log rules

x = \frac{1}{e ^{\frac{3 l n ( 2 )}{2}}}

Verify Solutions:

x = \frac{1}{e ^{\frac{3 l n ( 2 )}{2}}}

True

The solution is

x = \frac{1}{e ^{\frac{3 l n ( 2 )}{2}}}

ln (x) = 1 - ln (x - 3)

2 ln (x) - ln (x + 5) = ln (x) - ln (3)

2 ln (x) - ln (x - 1) = ln (4)

3.5 = 1.5 + lo g_{3} (x + 1)

lo g_{10} (3 x) = 4

What is the answer to ln(2x)+ln(4x)=0 ?
The answer to ln(2x)+ln(4x)=0 is x= 1/(e^{\frac{3ln(2)){2}}}
Solve for x: ln(2x)+ln(4x)=0
The solution is x= 1/(e^{\frac{3ln(2)){2}}}