Question

solve the following logarithmic equation, using a calculator if necessary to evaluate the logarithm. write your answer as a fraction or round your answer to two decimal places. (e ≈ 2.7183 )
ln(7x + 9) = 2
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Explanation:

Step1: Convert ln to exponential form

Recall that if $\ln(a) = b$, then $e^b = a$. So for $\ln(7x + 9) = 2$, we get $e^2 = 7x + 9$.

Step2: Solve for x

First, calculate $e^2$. Since $e \approx 2.7183$, $e^2 \approx 2.7183^2 \approx 7.3891$. Then the equation becomes $7.3891 = 7x + 9$. Subtract 9 from both sides: $7x = 7.3891 - 9 = -1.6109$. Then divide both sides by 7: $x = \frac{-1.6109}{7} \approx -0.23$? Wait, no, wait, I made a mistake. Wait, $e^2$ is about 7.389, but 7.389 - 9 is negative? Wait, no, the original equation is $\ln(7x + 9) = 2$. So $7x + 9 = e^2$. So $7x = e^2 - 9$. Let's compute $e^2 - 9$: $e^2 \approx 7.389056$, so $7.389056 - 9 = -1.610944$. Then $x = \frac{-1.610944}{7} \approx -0.23$? But that would make $7x + 9 = 7*(-0.23) + 9 = -1.61 + 9 = 7.39$, and $\ln(7.39) \approx 2$, which is correct. Wait, but let's do it properly.

Wait, step by step:

1. Start with $\ln(7x + 9) = 2$.
2. Exponentiate both sides with base $e$: $e^{\ln(7x + 9)} = e^2$.
3. Simplify left side: $7x + 9 = e^2$.
4. Subtract 9: $7x = e^2 - 9$.
5. Divide by 7: $x = \frac{e^2 - 9}{7}$.

Now compute $e^2 \approx 7.38905609893$, so $e^2 - 9 \approx 7.38905609893 - 9 = -1.61094390107$. Then $x \approx \frac{-1.61094390107}{7} \approx -0.23$? Wait, but that seems odd. Wait, maybe I messed up the sign? Wait, no, $\ln(7x + 9) = 2$ implies $7x + 9$ is positive, so $7x + 9 > 0 \implies x > -9/7 \approx -1.2857$. So $x \approx -0.23$ is greater than -1.2857, so it's valid. Wait, but let's check again.

Wait, $e^2$ is approximately 7.389, so $7x + 9 = 7.389 \implies 7x = 7.389 - 9 = -1.611 \implies x = -1.611 / 7 \approx -0.23$. Yes, that's correct.

Wait, but maybe I made a mistake in the sign. Wait, no, the calculation is correct. Let's verify: $7*(-0.23) + 9 = -1.61 + 9 = 7.39$. $\ln(7.39) \approx \ln(7.389) \approx 2$, since $e^2 \approx 7.389$. So that's correct.

Wait, but let's do it with more precise calculation. $e^2 = 7.3890560989306495$. So $e^2 - 9 = 7.3890560989306495 - 9 = -1.6109439010693505$. Then divide by 7: $x = -1.6109439010693505 / 7 \approx -0.2301348430099072$. Rounding to two decimal places, that's -0.23. Wait, but let's check if that's correct. Wait, maybe I made a mistake in the exponentiation. Wait, no, $\ln(a) = b \implies a = e^b$ is correct. So the steps are correct.

Wait, but maybe the problem is written incorrectly? Or maybe I misread it. Wait, the equation is $\ln(7x + 9) = 2$. So solving it gives $x = (e^2 - 9)/7 \approx (7.389 - 9)/7 \approx (-1.611)/7 \approx -0.23$.

Answer:

$\boxed{-0.23}$ (Wait, but let's check again. Wait, maybe I made a mistake in the sign. Wait, no, $e^2$ is about 7.389, so 7.389 - 9 is negative, so x is negative, but still greater than -9/7 ≈ -1.2857, so it's valid. So the answer is approximately -0.23.