Question

help pls nerds

Answer 1

d

Answer 2

is it actually?

Answer 3

yup

Answer 4

ugh im not typing it all out

Answer 5

well I would prefer to see the math done lol

Answer 6

damn

Answer 7

i used chat gpt

Answer 8

lol

Answer 9

ok I have more

Answer 10

oh no

Answer 11

Since no one else is going to, I'll just give you the general formular for change of base:

\[\large log_{a}(x) = \frac{log_{b}(x)}{log_{b}(a)} \]
Where b is the standard base of 10

So in your first problem, it would be \[\Large log_{13}(17.21) = \frac{log_{10}(17.21)}{log_{10}(13)} \]

Answer 12

so does that mean that the answer is c?

Answer 13

For the first one, it should be the same, so 1.109

Answer 14

yes that's what I thought I don't trust chat gpt because it isn't always accruate

Answer 15

For the second one, you're gonna use the general rules of:
\[\large ln(x)*ln(y) = ln(x)+ln(y) \]
\[\large ln (x) - ln (y) = ln \frac{x}{y} \]
\[\large a*ln (x)= x^a \]

Answer 16

so that means that i would get ln (13/w^1 1 x^5)?

Answer 17

This might be a long one so I'll try shortening it:
First condense the exponents

\[\large ln(13)+ln(x^3)+ln(w^{11})-ln(w^{12})-ln(x^8) \]
Now apply the multiplication/subtracting properties:

\[\large ln(13*x^3*w^{11})-ln(w^{12})-ln(x^8) \]
\[\large ln \frac{ (13*x^3*w^{11})}{(w^{12})} -ln(x^8) \]
\[\large ln \frac{ (13*x^3)}{(w)} -ln(x^8) \]
\[\large ln \frac{ (13*x^3)}{(w*x^8)}\]

\[\large ln \frac{ (13)}{(w*x^5)}\]

Answer 18

Might need to be double checked but that's the process

Answer 19

ln(13xw^11/x^8)

Answer 20

this is the answer I got

Answer 21

when I used the general rules you provided on the equation

Answer 22

oh wait I wrote something wrong

Answer 23

so its ln(13/w times x^8)?

Answer 24

*x^5 I'm pretty sure, cause of the \(\Large \frac{x^3}{x^8} \) reduces to \(\Large \frac{1}{x^5} \)

Answer 25

Ou nerds