Question

Log2(4x2+12x)=log2(2x2+6x)

Answer 1

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Answer 2

To solve the equation $\log_2(4x^2 + 12x) = \log_2(2x^2 + 6x)$, we can use properties of logarithms. The equation states that the logarithm base 2 of two expressions is equal, which implies that the expressions themselves are equal.

Here's the equation again for clarity: $4x^2 + 12x = 2x^2 + 6x$

Let's solve for $x$:

$4x^2 + 12x = 2x^2 + 6x$ $4x^2 - 2x^2 + 12x - 6x = 0$ $2x^2 + 6x = 0$ $2x(x + 3) = 0$

Now we have two possible solutions:

1. $2x = 0$ which gives $x = 0$
2. $x + 3 = 0$ which gives $x = -3$

So, the solutions to the equation are $x = 0$ and $x = -3$. You can plug these values back into the original equation to verify that they satisfy the equation.