Solve for the exact value of x. 3ln(2x-2)+14=5 Answer:

Isolate natural logarithm term: First, we need to isolate the natural logarithm term on one side of the equation. To do this, we subtract 14 from both sides of the equation. 3ln(2x-2) + 14 - 14 = 5 - 14 3ln(2x-2) = -9Divide by 3: Next, we divide both sides of the equation by 3 to solve for the natural logarithm of (2x-2). (3ln(2x-2))/3 = -9/3 ln(2x-2) = -3Exponentiate to remove ln: Now, we will exponentiate both sides of the equation to remove the natural logarithm. We use the property that e^(ln(x)) = x. e^(ln(2x-2)) = e^(-3) 2x - 2 = e^(-3)Add 2: We then add 2 to both sides of the equation to solve for 2x. 2x - 2 + 2 = e^(-3) + 2 2x = e^(-3) + 2Divide by 2: Finally, we divide both sides of the equation by 2 to solve for x. 2x/2 = (e^(-3) + 2)/2 x = (e^(-3) + 2)/2

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Q. Solve for the exact value of

x

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3 \ln (2 x-2)+14=5

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Answer:

Isolate natural logarithm term: First, we need to isolate the natural logarithm term on one side of the equation. To do this, we subtract $14$ from both sides of the equation. $\newline$ $3\ln(2x-2) + 14 - 14 = 5 - 14$ $\newline$ $3\ln(2x-2) = -9$
Divide by $3$ : Next, we divide both sides of the equation by $3$ to solve for the natural logarithm of $(2x-2)$ . $\frac{3\ln(2x-2)}{3} = \frac{-9}{3}$ $\ln(2x-2) = -3$
Exponentiate to remove ln: Now, we will exponentiate both sides of the equation to remove the natural logarithm. We use the property that $e^{\ln(x)} = x$ . $\newline$ $e^{\ln(2x-2)} = e^{-3}$ $\newline$ $2x - 2 = e^{-3}$
Add $2$ : We then add $2$ to both sides of the equation to solve for $2x$ . $\newline$ $2x - 2 + 2 = e^{-3} + 2$ $\newline$ $2x = e^{-3} + 2$
Divide by $2$ : Finally, we divide both sides of the equation by $2$ to solve for x. $\newline$ $\frac{2x}{2} = \frac{e^{-3} + 2}{2}$ $\newline$ $x = \frac{e^{-3} + 2}{2}$