# I have 4^(x+1) .. how to proceed

• March 24th 2008, 01:47 PM
Twig
I have 4^(x+1) .. how to proceed
Hi!

4^(x+1) + 3 * 4^x = 112

Should I use log or something?
• March 24th 2008, 02:02 PM
Jhevon
Quote:

Originally Posted by Twig
Hi!

4^(x+1) + 3 * 4^x = 112

Should I use log or something?

logarithms? no. but good thinking. you should be thinking logarithms for things like this, but in this case, logarithms are not necessary

note that $4^{x + 1} = 4 \cdot 4^x$

can you continue now?
• March 24th 2008, 02:08 PM
Twig
hi
I did solve it, but I doubt I did it "correctly".

I got 4 * 4^x + 3 * 4^x = 112

I divided all terms with 4^x

so I have 7 = 112/4^x

this gives me 4^x = 112/7

now I used logarithms to solve for x
• March 24th 2008, 02:10 PM
Jhevon
Quote:

Originally Posted by Twig
I did solve it, but I doubt I did it "correctly".

I got 4 * 4^x + 3 * 4^x = 112

I divided all terms with 4^x

so I have 7 = 112/4^x

this gives me 4^x = 112/7

now I used logarithms to solve for x

what is in red is completely unnecessary

now simplify first. 112/7 = 16

so you have $4^x = 16$

do you really need logarithms for that?
• March 24th 2008, 02:13 PM
Twig
hi

so I can basically say that 4 * 4^x + 3* 4^x = 7 * 4^x ?

and just from there take 112/7 = 16

and 4^x = 16
• March 24th 2008, 02:14 PM
Jhevon
Quote:

Originally Posted by Twig
hi

so I can basically say that 4 * 4^x + 3* 4^x = 7 * 4^x ?

and just from there take 112/7 = 16

and 4^x = 16

yes (you divide both sides by 7)

now what is the solution?
• March 24th 2008, 02:16 PM
Twig
x = 2
• March 24th 2008, 02:17 PM
Jhevon
Quote:

Originally Posted by Twig
x = 2

(Clapping) .....you didn't use logarithms, did you? :D

are we good now? do we understand the concept?
• March 24th 2008, 02:19 PM
Twig
thanks a lot
Yes I did.

I just didnt realize at first that I can add up my 4^x terms so I get 7 * 4^x
• March 24th 2008, 02:21 PM
Jhevon
Quote:

Originally Posted by Twig
thanks a lot
Yes I did.

I just didnt realize at first that I can add up my 4^x terms so I get 7 * 4^x

yes. you can essentially treat the 4^x as a variable. just as how if you had 4x + 3x you would get 7x. the idea is the same

or if you don't want to think of it that way, you can think of it as factoring out the 4^x
• March 24th 2008, 02:21 PM
Twig
alright cool =)