What is the answer to log_{8}(2)+log_{8}(4x^2)=1 ?

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

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log_{8}(2)+log_{8}(4x^2)=1

lo g_{8} (2) + lo g_{8} (4 x^{2}) = 1

x = 1, x = - 1

Solution steps

lo g_{8} (2) + lo g_{8} (4 x^{2}) = 1

Move

lo g_{8} (2)

to the right side

lo g_{8} (4 x^{2}) = 1 - lo g_{8} (2)

Apply log rules

4 x^{2} = 8^{1 - \frac{1}{3}}

Solve

4 x^{2} = 8^{1 - \frac{1}{3}} : x = 1, x = - 1

x = 1, x = - 1

Verify Solutions:

x = 1

True

, x = - 1

True

The solutions are

x = 1, x = - 1

lo g_{8} (x) = 3

lo g_{4} (x) + lo g_{x} (4) = 2.5

lo g_{10} (x) + lo g_{10} (8) = 2

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

lo g_{a} (8) = 3

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