What is the answer to log_{8}(3)+log_{8}(-4x)=1 ?

The answer to log_{8}(3)+log_{8}(-4x)=1 is x=-2/3

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log_{8}(3)+log_{8}(-4x)=1

lo g_{8} (3) + lo g_{8} (- 4 x) = 1

x = - \frac{2}{3}

Solution steps

lo g_{8} (3) + lo g_{8} (- 4 x) = 1

Move

lo g_{8} (3)

to the right side

lo g_{8} (- 4 x) = 1 - lo g_{8} (3)

Apply log rules

- 4 x = \frac{8}{3}

Solve

- 4 x = \frac{8}{3} : x = - \frac{2}{3}

x = - \frac{2}{3}

Verify Solutions:

x = - \frac{2}{3}

True

The solution is

x = - \frac{2}{3}

lo g_{6} (x + 1) + lo g_{6} (x + 2) = 1

lo g_{32} (m) = \frac{1}{5}

lo g_{2} (6 x + 3) = 5

lo g_{5} (x^{4}) + lo g_{5} (x) = 8

lo g_{10} (16) = 2 lo g_{10} (x + 1)

What is the answer to log_{8}(3)+log_{8}(-4x)=1 ?
The answer to log_{8}(3)+log_{8}(-4x)=1 is x=-2/3
Solve for x: log_{8}(3)+log_{8}(-4x)=1
The solution is x=-2/3