What is the answer to log_{2x}(3x^2+4x-3)=2 ?

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

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log_{2x}(3x^2+4x-3)=2

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

x = 1, x = 3

Solution steps

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

Apply log rules

\frac{ln ( 3 x ^{2} + 4 x - 3 )}{ln ( 2 x )} = 2

Multiply both sides by

ln (2 x)

\frac{ln ( 3 x ^{2} + 4 x - 3 )}{ln ( 2 x )} ln (2 x) = 2 ln (2 x)

Simplify

ln (3 x^{2} + 4 x - 3) = 2 ln (2 x)

Apply log rules

3 x^{2} + 4 x - 3 = 4 x^{2}

Solve

3 x^{2} + 4 x - 3 = 4 x^{2} : x = 1, x = 3

x = 1, x = 3

Verify Solutions:

x = 1

True

, x = 3

True

The solutions are

x = 1, x = 3

A, lo g_{ab} (A) = 6

lo g_{4} (12 x) = 3

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

lo g_{10} (x) = 4.1

lo g_{2} (x) + lo g_{4} (x) + lo g_{16} (x) = 7

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