Solve for the exact value of x. 4ln(7x+3)+19=43 Answer:

Isolate logarithmic expression: First, we need to isolate the logarithmic expression by subtracting 19 from both sides of the equation. 4ln(7x+3) + 19 - 19 = 43 - 19 4ln(7x+3) = 24Divide by 4: Next, we divide both sides by 4 to solve for the natural logarithm of (7x+3). (4ln(7x+3)) / 4 = 24 / 4 ln(7x+3) = 6Exponentiate both sides: Now, we will exponentiate both sides to remove the natural logarithm, using the property e^(ln(x)) = x. e^(ln(7x+3)) = e^6 7x + 3 = e^6Subtract 3: Subtract 3 from both sides to isolate the term with x. 7x + 3 - 3 = e^6 - 3 7x = e^6 - 3Divide by 7: Finally, divide both sides by 7 to solve for x. 7x / 7 = (e^6 - 3) / 7 x = (e^6 - 3) / 7

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Solve for the exact value of
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4ln(7x+3)+19=43
Answer:

Solve for the exact value of $x$ . $\newline$ $4 \ln (7 x+3)+19=43$ $\newline$ Answer:

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Algebra 1

Solve linear equations: mixed review

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Q. Solve for the exact value of

x

\newline

4 \ln (7 x+3)+19=43

\newline

Answer:

Isolate logarithmic expression: First, we need to isolate the logarithmic expression by subtracting $19$ from both sides of the equation. $\newline$ $4\ln(7x+3) + 19 - 19 = 43 - 19$ $\newline$ $4\ln(7x+3) = 24$
Divide by $4$ : Next, we divide both sides by $4$ to solve for the natural logarithm of $(7x+3)$ . $\frac{4\ln(7x+3)}{4} = \frac{24}{4}$ $\ln(7x+3) = 6$
Exponentiate both sides: Now, we will exponentiate both sides to remove the natural logarithm, using the property $e^{\ln(x)} = x$ . $\newline$ $e^{\ln(7x+3)} = e^6$ $\newline$ $7x + 3 = e^6$
Subtract $3$ : Subtract $3$ from both sides to isolate the term with $x$ . $\newline$ $7x + 3 - 3 = e^{6} - 3$ $\newline$ $7x = e^{6} - 3$
Divide by $7$ : Finally, divide both sides by $7$ to solve for x. $\newline$ $\frac{7x}{7} = \frac{e^6 - 3}{7}$ $\newline$ $x = \frac{e^6 - 3}{7}$