What is the answer to log_{2x}(2x^2+6x-4)=2 ?

The answer to log_{2x}(2x^2+6x-4)=2 is x=1,x=2

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log_{2x}(2x^2+6x-4)=2

lo g_{2 x} (2 x^{2} + 6 x - 4) = 2

x = 1, x = 2

Solution steps

lo g_{2 x} (2 x^{2} + 6 x - 4) = 2

Apply log rules

\frac{ln ( 2 x ^{2} + 6 x - 4 )}{ln ( 2 x )} = 2

Multiply both sides by

ln (2 x)

\frac{ln ( 2 x ^{2} + 6 x - 4 )}{ln ( 2 x )} ln (2 x) = 2 ln (2 x)

Simplify

ln (2 x^{2} + 6 x - 4) = 2 ln (2 x)

Apply log rules

2 x^{2} + 6 x - 4 = 4 x^{2}

Solve

2 x^{2} + 6 x - 4 = 4 x^{2} : x = 1, x = 2

x = 1, x = 2

Verify Solutions:

x = 1

True

, x = 2

True

The solutions are

x = 1, x = 2

10 lo g_{10} (x^{2}) = - 20

2 5^{l o g_{5} (3 x^{2})} = 9 x^{2}

lo g_{10} (x) = - 13.9

lo g_{2} (x + 19) + lo g_{2} (x + 4) = 4

x^{1 + l n (x)} = e^{4} \cdot x

What is the answer to log_{2x}(2x^2+6x-4)=2 ?
The answer to log_{2x}(2x^2+6x-4)=2 is x=1,x=2
Solve for x: log_{2x}(2x^2+6x-4)=2
The solution is x=1,x=2