Solve for the exact value of x. 3ln(7x-6)+19=31 Answer:

Isolate natural logarithm: First, we need to isolate the natural logarithm term on one side of the equation. 3ln(7x-6) + 19 = 31 Subtract 19 from both sides to get the natural logarithm by itself. 3ln(7x-6) = 31 - 19 3ln(7x-6) = 12Divide by 3: Next, we divide both sides by 3 to solve for the natural logarithm of (7x-6). (3ln(7x-6)) / 3 = 12 / 3 ln(7x-6) = 4Exponentiate both sides: Now, we will exponentiate both sides to remove the natural logarithm and solve for the expression inside it. e^(ln(7x-6)) = e^4 7x - 6 = e^4Add 6: We then add 6 to both sides to isolate the term with x. 7x - 6 + 6 = e^4 + 6 7x = e^4 + 6Divide by 7: Finally, we divide both sides by 7 to solve for x. 7x / 7 = (e^4 + 6) / 7 x = (e^4 + 6) / 7

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Solve for the exact value of
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3ln(7x-6)+19=31
Answer:

Solve for the exact value of $x$ . $\newline$ $3 \ln (7 x-6)+19=31$ $\newline$ Answer:

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Solve linear equations: mixed review

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Q. Solve for the exact value of

x

\newline

3 \ln (7 x-6)+19=31

\newline

Answer:

Isolate natural logarithm: First, we need to isolate the natural logarithm term on one side of the equation. $\newline$ $3\ln(7x-6) + 19 = 31$ $\newline$ Subtract $19$ from both sides to get the natural logarithm by itself. $\newline$ $3\ln(7x-6) = 31 - 19$ $\newline$ $3\ln(7x-6) = 12$
Divide by $3$ : Next, we divide both sides by $3$ to solve for the natural logarithm of $(7x-6)$ . $\frac{3\ln(7x-6)}{3} = \frac{12}{3}$ $\ln(7x-6) = 4$
Exponentiate both sides: Now, we will exponentiate both sides to remove the natural logarithm and solve for the expression inside it. $\newline$ $e^{\ln(7x-6)} = e^4$ $\newline$ $7x - 6 = e^4$
Add $6$ : We then add $6$ to both sides to isolate the term with $x$ . $\newline$ $7x - 6 + 6 = e^4 + 6$ $\newline$ $7x = e^4 + 6$
Divide by $7$ : Finally, we divide both sides by $7$ to solve for x. $\newline$ $\frac{7x}{7} = \frac{e^4 + 6}{7}$ $\newline$ $x = \frac{e^4 + 6}{7}$