Question

16^(6x) * (1/4)^(9x-5) = 64 * (4^x)^-7 Solve

Answer 1

are you kidding?

Answer 2

Why would I be?

Answer 3

gimmick is to rearrange this so that both sides are powers of 4. you get
\[4^{3x-7}\] on the left

Answer 4

sorry. that is the right hand side, not the left hand side

Answer 5

right hand side is
\[4^{3x-7}\]

Answer 6

left hand side is
\[4^{3x+5}\]

Answer 7

once that algebra nightmare is over it is pretty simple. since they are both powers of 4 you can just set the exponents equal and solve:
\[3x+5=3-7x\]
\[10x=-2\]
\[x=\frac{-2}{10}=-\frac{1}{5}\]

Answer 8

oh lord i see i made a typo in earlier post sorry

Answer 9

right hand side is
\[4^{3-7x}\] not what i wrote

Answer 10

answer is correct though

Answer 11

No problem, thanks! I see where I went wrong XD

Answer 12

hope you see where i went wrong too. but answer is correct.

Answer 13

\[\color{blue}{\text{hello my myininaya!}}\]

Answer 14

im sorry but are you sure you wrote the problem down correctly autumn

Answer 15

\[\color{#FF0018}{\text{ are you in the pink?}}\]

Answer 16

\[\color{red}{\text{don't cause trouble}}\]

Answer 17

hmmm maybe i'm wrong i read the last thingy as x^(-7)

Answer 18

\[16^{6x} \times( \frac{1}{4})^{9x-5} = 64 \times (4^x)^{-7}\]

Answer 19

okay that looks prettier
:)

Answer 20

i accept the answer as -.2 now

Answer 21

as i said my little \[\color{red}{\text{apple pi}}\]

Answer 22

lol

Answer 23

right hand side is
\[4^{7-3x}\]

Answer 24

ok maybe it is is
\[4^{3-7x}\]

Answer 25

\[\color{blue}{\text{typo!}}\]

Answer 26

it is the second one

Answer 27

what kind of chump math teacher would give this problem?

Answer 28

stupid gimmick of writing everything in the same base. sheesh