Question

solve for x

Answer 1

\[\frac{ 1 }{ 9 }^x=27^2x+4\]

Answer 2

27^2x+4 is all one

Answer 3

What do you mean is all one?

Can't you simplify 27^2 already? Mhmm

Answer 4

no it's an exponential function

Answer 5

and 27 is to the power of 2x+4

Answer 6

ohhh

Answer 7

yeah sorry i screwed it up

Answer 8

it's alright, not everyone knows how to use latex :P

\(\Large \frac{1}{9}^x = 27^{2x+4}\)

we need to get both sides to have the same base so we can use this

\( a^c = a^d \rightarrow c = d\)

Answer 9

k that's where I get lost

Answer 10

like I don't now how to make them have the same base

Answer 11

I know it has something to do with exponents tho

Answer 12

rewrite 9 to have a base 3
rewrite 27 to have a base 3

3^x = 9
what is x?

3^x = 27
x = ?

Answer 13

so the fact that it's a fraction doesn't change what yo do to make the bases the same?

Answer 14

x=2 x=3/1

Answer 15

i think I'm not sure about the second one

Answer 16

no, we'll be using the property

a^(-b) = 1/a^b

you're correct on both

Answer 17

okay cool

Answer 18

now we can change \(\Large(\frac{1}{9})^x= (\frac{1}{3^2})^x =(3^{-2})^x\)

do you understand? :)

Answer 19

ohhh yeah I think it was the negatives that threw me off like I didn't know why some of the exponents were negative

Answer 20

and we replace 27 with 3^3

\(\Large 3^{-2x} = 27^{2x+4}\)

\(\Large 3^{-2x} = (3^3)^{2x+4}\)

now we need to apply the rule \(\Large (a^b)^c = a^{bc}\)

you'll need to distribute the 3 with 2x + 3

what do you get as the exponent on the right hand side?

Answer 21

6x+12

Answer 22

now we have

\(\Large 3^{-2x} = 3^{6x+12}\)

remember this rule?

If \(\Large a^b = a^c\) then \(\large b = c\)

so all you need to solve now is

-2x = 6x + 12

solve for x :)

Answer 23

x=-3

Answer 24

oops wait i messed up

Answer 25

not quite

subtract 6x on both sides
and isolate x

Answer 26

sure, I'll wait :)

Answer 27

lol sorry I got x=-3/2

Answer 28

great work :D

Answer 29

yaya thank you!

Answer 30

Anytime! :D