4^2x - 17(4^x) + 16 = 0
Find x.

Thanks in advance.

Let Y=4^x

Y^2 - 17Y + 16 = 0

(Y-16) (Y-1) = 0

Y = 16 OR Y = 1

4^x = 16 OR 4^x = 1

You should be able to work it out easily at this point but if not:

log[SIZE="1"]4[/SIZE] 16 = x = 2

OR

log[SIZE="1"]4[/SIZE] 1 = x = 1 Edit: correction. log[SIZE="1"]4[/SIZE] 1 = 0

x = 2 OR 0

(unless of course you're working in graphical form and trying to find points of intersection in which case x = 2 AND 0)

Now gimme a hard one!

Log3^25 divided by Log3^5

Thanks man.

Pretty sure that log3^25 / log3^5 = log3^20

I hope you haven't handed in that question yet. I just realised I made a simple arithmetic error at the end - log of 1 is 0, not 1. X=2 or 0.

Steyene looks right on the other one.

oh my word u guys r good!

:Happy: lol:Smile: :Smile: :Smile:

xD so randum

1+1=11

My good deed for the day is done.