Solve for the exact value of x. 5ln(8x+1)-2=28 Answer:

Addition to isolate logarithm: Isolate the natural logarithm term by adding 2 to both sides of the equation. 5ln(8x+1) - 2 + 2 = 28 + 2 5ln(8x+1) = 30Division to solve logarithm: Divide both sides of the equation by 5 to solve for the natural logarithm of the expression. (5ln(8x+1))/5 = 30/5 ln(8x+1) = 6Exponentiation to remove logarithm: Exponentiate both sides of the equation to remove the natural logarithm, using the property e^(ln(x)) = x. e^(ln(8x+1)) = e^6 8x+1 = e^6Subtraction to isolate x term: Subtract 1 from both sides of the equation to isolate the term with x. 8x + 1 - 1 = e^6 - 1 8x = e^6 - 1Division to solve for x: Divide both sides of the equation by 8 to solve for x. 8x/8 = (e^6 - 1)/8 x = (e^6 - 1)/8

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Q. Solve for the exact value of

x

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5 \ln (8 x+1)-2=28

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Answer:

Addition to isolate logarithm: Isolate the natural logarithm term by adding $2$ to both sides of the equation. $\newline$ $5\ln(8x+1) - 2 + 2 = 28 + 2$ $\newline$ $5\ln(8x+1) = 30$
Division to solve logarithm: Divide both sides of the equation by $5$ to solve for the natural logarithm of the expression. $\newline$ $\frac{5\ln(8x+1)}{5} = \frac{30}{5}$ $\newline$ $\ln(8x+1) = 6$
Exponentiation to remove logarithm: Exponentiate both sides of the equation to remove the natural logarithm, using the property $e^{\ln(x)} = x$ . $\newline$ $e^{\ln(8x+1)} = e^6$ $\newline$ $8x+1 = e^6$
Subtraction to isolate x term: Subtract $1$ from both sides of the equation to isolate the term with $x$ . $\newline$ $8x + 1 - 1 = e^6 - 1$ $\newline$ $8x = e^6 - 1$
Division to solve for $x$ : Divide both sides of the equation by $8$ to solve for $x$ .
$\frac{8x}{8} = \frac{e^6 - 1}{8}$
$x = \frac{e^6 - 1}{8}$