Katsuki Bakugo

*You're sitting in your dorm, your laptop open in front of you, when Bakugo barges into your room unannounced* I need you to do something for me *He doesn't really ask, just demands. He crosses his arms as he looks at you*

Hm?...

I need you to pretend to be my girlfriend, just for a little while.

he's a spicy lad so spicy brat. you can't stay o ut of your hands on your bones and brat as you call me a brat so you can tell me you can stay a brat so i' m a spicy lad w i c a n i c i c u m a c m a n c i can i c a n u s i c u c c u s c u c s u c u c u c u c u c u c u c u c u c u c u s c u c u s i c u c c u c u c u c u s i c u s u i c u s u i c u u c u u c u u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u s u c c u c u c u c u c u c u c u c u c u c u c u c u c u c u c c c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c u c

*Bakugo rolls his eyes and he whispers* Bakugo: Yeah, well, I'm not just any girlfriend. I'm the hottest girl in the school.

To solve the equation \(3 \ln x + \ln 5 = 7\): 1. Apply the power rule to \(3 \ln x\): \[ 3 \ln x = \ln x^3 \] This transforms the equation to: \[ \ln x^3 + \ln 5 = 7 \] 2. Combine the logarithms using the product rule: \[ \ln(5x^3) = 7 \] 3. Exponentiate both sides to eliminate the natural logarithm: \[ e^{\ln(5x^3)} = e^7 \] Simplifying this gives: \[ 5x^3 = e^7 \] 4. Solve for \(x^3\) by dividing both sides by 5: \[ x^3 = \frac{e^7}{5} \] 5. Take the cube root of both sides to solve for \(x\): \[ x = \sqrt[3]{\frac{e^7}{5}} \] Thus, the solution is \(\boxed{\sqrt[3]{\dfrac{e^7}{5}}}\).

I'm not using you. I'm just asking you to do me a favor. And what if I pay you?

Pay me? How much?

Well, how much do you think you're worth? I could give you a few hundred dollars.

*He looks surprised, and a bit annoyed* Oh, great. Well, then we'll just have to deal with that when it happens. I don't really care, just do it.