How do I solve for x when 3 ln(2x+1) = 9?

[Paisley_W]

I keep getting stuck when I try to solve for a variable inside a natural logarithm, like in the equation 3 ln(2x+1) = 9. I understand the basic log rules, but isolating 'x' here feels like I'm missing a clear step in the process.

[LarryH]

Yeah I get that. I’d undo the ln by dividing first: ln(2x+1) = 3. Then exponentiate both sides: 2x+1 = e^3, so x = (e^3 - 1)/2. Domain wise you need 2x+1 > 0, so x > -1/2, which keeps that solution valid. I’m not totally sure I didn’t miss a sign somewhere, but that’s the path I’d try.

[Zoey68]

I plugged it into a calculator and got x around 9.54. Because e^3 is about 20.085, subtract 1 is 19.085, divide by 2 is 9.5425. It feels big for a simple log, but that’s what the numbers give.

[Anthony_P]

I remember the trick that you can rewrite the left side as a cube inside the log, then exponentiate. Is it okay to just exponentiate and skip straight to 2x+1 = e^3? The domain would still force 2x+1 > 0, so it lines up with x > -1/2.

[Paul86]

I’ve spent a while on a similar problem and ended up chasing a different issue, like whether I actually picked the right side to solve. Sometimes stepping back and sketching a quick graph helps me notice I’m solving for the wrong thing, not the math itself.