Question

help

Answer 1

@Vocaloid

Answer 2

@Ultrilliam

Answer 3

@hero @angle I've been having really bad migraines today can you assist with some math?

Answer 4

help

Answer 5

@kaylak the picture is really blurry for me (or maybe I need new glasses) could you upload a clearer one?
... or post one question at a time?

Answer 6

sure

Answer 7

ooo much better
now...
Have you seen this exponent rule before?
\[\huge a^{b - c} = \frac{a^b}{a^c}\]

Answer 8

no lol

Answer 9

typing a lot lol

Answer 10

ok, for example

\(\huge \frac{2^6}{2^2} = \frac{2*2*2*2*2*2}{2*2}\)

\(\large = 2*2*2*2 = 2^4\)

yes?

Answer 11

yes

Answer 12

which means that

\(\huge \frac{2^6}{2^2} \large = 2^{6-2} = 2^4\)

which is basically the rule I stated above

does that make sense?

Answer 13

yes

Answer 14

d then?

Answer 15

so

\(\large a^{b-c} = \huge \frac{a^b}{a^c}\)

Answer 16

if -3 stays -3 and 4^(1-x) turns into 1/4^x

Answer 17

right

Answer 18

?

Answer 19

That's the think I was trying to say by starting with stating that rule I was talking about...

\(\large 4^{1-x} = \huge \frac{4^1}{4^x}\)

Answer 20

so is d the answer

Answer 21

no
because 4^(1-x) is *not* 1/(4^x)

Answer 22

oh ok it's a then so you have to multiply 3*4

Answer 23

oh ok it's a then so you have to multiply 3*4

Answer 24

yup!

Answer 25

is 2 correct

Answer 26

2 being d

Answer 27

?

Answer 28

\(\color{#0cbb34}{\text{Originally Posted by}}\) @kaylak

\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 29

can't quote a file

Answer 30

hmmm that's not what I got...

you basically want to solve for

(x -1)(2x + 3) = 1

Answer 31

that's what mathway got?

Answer 32

nope... wolframalpha...

but the idea is that for logs of the same base:
\(\large ~log_a x\ + log_a y = log_a (x*y) \)

which is how you end up with trying to solve for

(x - 1)(2x + 3) = 1

Answer 33

ohhh wait

you did get the right idea

BUT
you cannot have the log(negative number)

Answer 34

so c then?

Answer 35

yeah ^_^

Answer 36

lol next 2

Answer 37

I need to go to class now
feel free to ask your next two questions as a new post for someone else to help you ^_^