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Find all values of $x$ . $\newline$ $\log_{2}(2x^{2}+2)-\log_{2}(3x+1)=0$

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Product property of logarithms

Full solution

Q. Find all values of

x

.

\newline

\log_{2}(2x^{2}+2)-\log_{2}(3x+1)=0

Combine logarithms: Use the property of logarithms to combine the two logarithms into a single logarithm. The subtraction of logarithms with the same base can be rewritten as the logarithm of the quotient of the arguments. $\log_2(2x^2+2) - \log_2(3x+1) = \log_2\left(\frac{2x^2+2}{3x+1}\right)$
Set equal and solve: Set the combined logarithm equal to zero and solve for the argument. $\newline$ Since $\log_2\left(\frac{2x^2+2}{3x+1}\right) = 0$ , we can use the definition of a logarithm to write this as an exponential equation. $\newline$ $2^{0} = \frac{2x^2+2}{3x+1}$
Simplify exponential equation: Simplify the exponential equation. $2^{0}$ is equal to $1$ , so we have: $1 = \frac{2x^{2}+2}{3x+1}$
Clear denominator: Multiply both sides of the equation by $(3x+1)$ to clear the denominator. $\newline$ $1 \times (3x+1) = (2x^2+2)$ $\newline$ $3x + 1 = 2x^2 + 2$
Rearrange to quadratic: Rearrange the equation to form a quadratic equation. $\newline$ $0 = 2x^2 - 3x + (2 - 1)$ $\newline$ $0 = 2x^2 - 3x + 1$
Factor or use formula: Factor the quadratic equation, if possible, or use the quadratic formula to find the values of $x$ . The quadratic equation $2x^2 - 3x + 1$ can be factored as $(2x - 1)(x - 1) = 0$
Solve for x: Set each factor equal to zero and solve for x. $\newline$ $2x - 1 = 0$ or $x - 1 = 0$ $\newline$ For $2x - 1 = 0$ , $x = \frac{1}{2}$ $\newline$ For $x - 1 = 0$ , $x = 1$

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