Solve for the exact value of x. 6ln(7x+1)+8=-10 Answer:

Isolate natural logarithm term: Isolate the natural logarithm term. We start by subtracting 8 from both sides of the equation to isolate the term with the natural logarithm. 6ln(7x+1) + 8 - 8 = -10 - 8 6ln(7x+1) = -18Divide to solve natural logarithm: Divide both sides by 6 to solve for the natural logarithm of the expression. 6ln(7x+1) / 6 = -18 / 6 ln(7x+1) = -3Exponentiate to remove logarithm: Exponentiate both sides to remove the natural logarithm. e^(ln(7x+1)) = e^(-3) 7x + 1 = e^(-3)Subtract to solve for 7x: Subtract 1 from both sides to solve for 7x. 7x + 1 - 1 = e^(-3) - 1 7x = e^(-3) - 1Divide to solve for x: Divide both sides by 7 to solve for x. 7x / 7 = (e^(-3) - 1) / 7 x = (e^(-3) - 1) / 7

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Solve for the exact value of
x.

6ln(7x+1)+8=-10
Answer:

Solve for the exact value of $x$ . $\newline$ $6 \ln (7 x+1)+8=-10$ $\newline$ Answer:

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Q. Solve for the exact value of

x

\newline

6 \ln (7 x+1)+8=-10

\newline

Answer:

Isolate natural logarithm term: Isolate the natural logarithm term. $\newline$ We start by subtracting $8$ from both sides of the equation to isolate the term with the natural logarithm. $\newline$ $6\ln(7x+1) + 8 - 8 = -10 - 8$ $\newline$ $6\ln(7x+1) = -18$
Divide to solve natural logarithm: Divide both sides by $6$ to solve for the natural logarithm of the expression. $\newline$ $\frac{6\ln(7x+1)}{6} = \frac{-18}{6}$ $\newline$ $\ln(7x+1) = -3$
Exponentiate to remove logarithm: Exponentiate both sides to remove the natural logarithm. $\newline$ $e^{\ln(7x+1)} = e^{-3}$ $\newline$ $7x + 1 = e^{-3}$
Subtract to solve for $7x$ : Subtract $1$ from both sides to solve for $7x$ . $\newline$ $7x + 1 - 1 = e^{-3} - 1$ $\newline$ $7x = e^{-3} - 1$
Divide to solve for x: Divide both sides by $7$ to solve for $x$ . $\frac{7x}{7} = \frac{e^{(-3)} - 1}{7}$ $x = \frac{e^{(-3)} - 1}{7}$