Question

64^(2-x)=4^(4x)... can someone please help me and explain how to solve this? thank you :)

Answer 1

use the fact:\[64=4^3\]

Answer 2

the rest is just loggy

Answer 3

you don't need to use logs

Answer 4

you do if you need the practice lol

Answer 5

you should end up with 4^{something} = 4^{something-else}
then just equate the powers

Answer 6

equate exponents is the same as logging

Answer 7

when u smarties are done here, come my way! haha

Answer 8

sunshine - do you understand how to solve this now?

Answer 9

no wait I am confused :/

Answer 10

i am trying to find the solution but idk how.

Answer 11

divide each side by 4^3 might plausible

Answer 12

you are given this:\[64^{2-x}=4^{4x}\]now replace the 64 with \(4^3\) to get:\[(4^3)^{2-x}=4^{4x}\]therefore:\[4^{3(2-x)}=4^{4x}\]

Answer 13

yep, then ignore the 4s and equate the exponential bits

Answer 14

Oh!And then what would I have to do?

Answer 15

Ignore all 4s?

Answer 16

if you are given:\[4^a=4^b\]what does that tell you about a and b?

Answer 17

lol, it makes no sense to ignore ALL the 4s, just the useless ones ;)

Answer 18

sunshine: can you answer my previous question above?

Answer 19

Does it mean they're the same? I am sorry I am not positive

Answer 20

\[log_4[4^a=4^b]\]
\[log_4(4^a)=log_4(4^b)\]
\[a\ log_4(4)=b\ log_4(4)\]
\[\frac{a\ \cancel{log_4(4)}}{\cancel{log_4(4)}}=\frac{b\ \cancel{log_4(4)}}{\cancel{log_4(4)}}\]
\[a=b\]

Answer 21

yes - thats right - it means a=b
so use this fact to work out the solution to your problem.

Answer 22

remember we ended up with:\[4^{3(2-x)}=4^{4x}\]

Answer 23

Alright, Thank you!!!

Answer 24

yw

Answer 25

IS the answer 6/7?

Answer 26

yup - well done!