Question

Hey, can someone please help me solve 4^2x - 5 x 4^x + 5 = 0 ?

Answer 1

\[4^{2x} - 5\times4^{x} + 5 = 0\] - I tried updating the orignal post but it didn't work so here is the equation

Answer 2

4^x = t

Answer 3

helps ?

Answer 4

4^2x - 5 x 4^x = 0
4^2x = 5 x 4^x
take logs of both sides:
ln 4^2x = ln 5 * 4^x

Answer 5

Is it possible to do it without logarithms? We have not yet done logarithms in class so it may give me bonus marks for answering using logarithms I don't think that is how it is meant to be solved

Answer 6

you have to substitute 4^x = t
then your equation will become:
t^2 -5t +5 = 0

Answer 7

Ah okay, that seems more like what we are doing, thanks for the help :)

Answer 8

now if indeed t will be something nice
youll be able to find x without using logs

Answer 9

So then just factorise and solve that - I will need to use quadratic formula though won't I because I can not see anyway to get that factorised

Answer 10

yes

Answer 11

Cool, thanks for the help :)

Answer 12

i guess youll have to use logs
since you wont get nice numbers.

Answer 13

My bad, I typed the question wrong accidently. It is plus 4 not plus 5 so therefore it does give me good numbers :P

Answer 14

Becomes (t-4)(t-1)

Answer 15

oh so solutions are obvious :)

Answer 16

Yep :P I had the question right the first time and when I tried to update using the formula thing I changed to a 5 instead of a 4 by mistake