help

[signal heath]

How to solve it? Please describe it too

[vapid bobcat]

You know 2^a=3^b but 16=2^4

Then use the properties of powers

[bright mural]

...no it doesn't, unless a and b aren't integers.

Wait, shit, nevermind, sorry.

[vapid bobcat]

Np, I was starting to panic lol

[bright mural]

Though, of course a and b aren't integers.

[signal heath]

Can u solve it please?

[vapid bobcat]

(16)^a/b=(2^4)^a/b =(2^a)^4/b

You know 2^a=3^b…

[signal heath]

Yup

[vapid bobcat]

So replace 2^a by 3^b the b s cancel out, you then get a result

[signal heath]

I still didn't get it , i unfortunately forgot this one 😭

[vapid bobcat]

You get (3^b)^4/b =3^(4b/b)

It’s using the properties of powers

The result is then 3^4

[signal heath]

Okay then?

Ow okay thanks a lot

[signal heath]

I didn't understand this line but thanks a lot for your concern 😊

[odd cypress]

@signal heath has given 1 rep to @vapid bobcat

[vapid bobcat]

Np, it’s just the properties of exponants

Like (a)^bc=(a^b)^c

[signal heath]

Owww it's the formula right?

[vapid bobcat]

Yes that’s right

[signal heath]

Okayyyy thank youuu soo much...Now I get it thanks thanks

[odd cypress]

@signal heath has given 1 rep to @vapid bobcat