Can someone please explain for me. (2-3i)^2

Question

oaktree · Answer

Sure. So we should write this out first, and we get$$(2-3i)(2-3i) = 4 - 6i - 6i +9i^2$$Following so far?

oaktree · Answer

@axelneel are you there?

anonymous · Answer

yes sorry i had another tab open. But that makes much more sense. why could wen not just use the rule (mn)^x=m^xn^x?

anonymous · Answer

it it because there is an addition symbol

oaktree · Answer

Yes. They are two different terms, not one.

anonymous · Answer

i see. what makes multiplying two terms okay to distribute the exponent to but not addition?

oaktree · Answer

Just think about it like they were two numbers. So$$(2*4)^2 = (8)^2 = 64$$But $$(2+4)^2 = 6^2 = 36$$They're totally different numbers on the inside.

anonymous · Answer

i see because of law of mutlication. 2^2*4^2=64

oaktree · Answer

Yeah.

anonymous · Answer

can you help me to simplfy -logx-logy+2logz into a single logorithm. where would i start?

oaktree · Answer

So we have$$-(\log x + \log y) + 2\log z = -(\log xy) + 2 \log z = \log z^2 - \log xy$$Can you continue from there? There's just one step left.

oaktree · Answer

Hint: $$\log a - \log b = \log \frac{ a }{ b }$$

anonymous · Answer

i seee, thank you very much

anonymous · Answer

would it be logz^2/logxy?