Question

Log/Expo Question: Solve 4^5-3x=1/64^1-x

Answer 1

\[4^{5-3x}=\frac{ 1 }{ 64^{1-x} }\]

Answer 2

start with
\[4^{5-2x}=64^{-(1-x)}\]

Answer 3

then go to
\[4^{5-2x}=64^{x-1}\] and since \(64=4^2\) you can write the right hand side as \((4^2)^{x-1}=4^{2x-2}\) by the laws of exponents

Answer 4

now you have
\[4^{5-3x}=4^{2x-2}\] so the exponents must be the same
set them equal and solve

Answer 5

64= 4^3

Answer 6

oh well

Answer 7

caught napping again
no matter, we can adjust

Answer 8

im trying to understand, i promise. Lol, just a sec

Answer 9

\(\huge 4^{5-3x}=4^{3x-3}\)

Answer 10

i made a mistake, fortunately @hartnn caught it

Answer 11

the basic idea is to write both sides with the same base, then equate the exponents

Answer 12

When you flipped the fraction in the beginning when did you go from negative back to positive?

Answer 13

Nevermind! lol i see..just a sec :)

Answer 14

I UNDERSTANDD!!! Yayyy, lol thank you both @satellite73 & @hartnn :)

Answer 15

good :)
welcome.

Answer 16

wait... so is x=6/8?

Answer 17

4/3*

Answer 18

yes, which can be simplified to ?

Answer 19

3/4* lol

Answer 20

Thanks :)

Answer 21

correct, finally.....

Answer 22

haha I know right :( dont judge me.. im tired :P

Answer 23

i won't , u did good work! understanding things.....

Answer 24

Thanks :) I will probably be asking more soon.. lol.

Answer 25

sure :)

Answer 26

@hartnn Can I ask you something really quick.. its a question similar to this one.. its..\[8^x=\frac{ 1 }{ 2^{\frac{ -3 }{ x}} }\]

Ok, now when I am flipping the fraction does it become \[8^x=(2^3)^\frac{ 3 }{ x }\] ??