What is the answer to log_{3}(x+4)=log_{3}(1-x) ?

The answer to log_{3}(x+4)=log_{3}(1-x) is x=-3/2

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log_{3}(x+4)=log_{3}(1-x)

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

x = - \frac{3}{2}

Solution steps

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

Apply log rules

x + 4 = 1 - x

Solve

x + 4 = 1 - x : x = - \frac{3}{2}

x = - \frac{3}{2}

Verify Solutions:

x = - \frac{3}{2}

True

The solution is

x = - \frac{3}{2}

lo g_{x} (\frac{1}{6}) = - 1

lo g_{7} (6 - 2 t) = 2

ln (ln (x)) = 3

lo g_{2} (\frac{3 x + 1}{4 x - 3}) = 3

lo g_{x} (6) = \frac{1}{2}

What is the answer to log_{3}(x+4)=log_{3}(1-x) ?
The answer to log_{3}(x+4)=log_{3}(1-x) is x=-3/2
Solve for x: log_{3}(x+4)=log_{3}(1-x)
The solution is x=-3/2