What is the answer to 2log_{e}(x+3)=log_{e}(9x+19) ?

The answer to 2log_{e}(x+3)=log_{e}(9x+19) is x=5,x=-2

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2log_{e}(x+3)=log_{e}(9x+19)

2 lo g_{e} (x + 3) = lo g_{e} (9 x + 19)

x = 5, x = - 2

Solution steps

2 lo g_{e} (x + 3) = lo g_{e} (9 x + 19)

Apply log rules

(x + 3)^{2} = 9 x + 19

Solve

(x + 3)^{2} = 9 x + 19 : x = 5, x = - 2

x = 5, x = - 2

Verify Solutions:

x = 5

True

, x = - 2

True

The solutions are

x = 5, x = - 2

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

lo g_{4} (x + 3) = lo g_{4} (9)

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

lo g_{5} (4 x - 3) = 1

lo g_{5} (4 x - 3) = 2

What is the answer to 2log_{e}(x+3)=log_{e}(9x+19) ?
The answer to 2log_{e}(x+3)=log_{e}(9x+19) is x=5,x=-2