Question

Find x from the equation: 4^x + 2^x = 12.

I GIVE MEDALS. PLEASE HELP ME :D

Answer 1

let z = 2^x

square both sides to get

z^2 = (2^x)^2

z^2 = 2^(2x)

z^2 = (2^2)^x

z^2 = 4^x

Answer 2

So because z = 2^x and z^2 = 4^x, we go from

4^x + 2^x = 12

to

z^2 + z = 12

Answer 3

Your job from here is to solve z^2 + z = 12 for z

You'll get two solutions for z.

Once you get those two solutions, use each one to find the solutions in terms of x.

Answer 4

Tell me what you get when you solve z^2 + z = 12 for z

Answer 5

@jim_thompson5910 i got z = 3 and z = -4 :)

Answer 6

@jim_thompson5910 so, i will try, 3 = 2^x and -4 = 2^x right??

Answer 7

very good

Answer 8

if 3 = 2^x, then x = ???

Answer 9

@jim_thompson5910 3 = 2^x.. can it be solved without a scientific calculator or anything like logarithm or something?? :)

Answer 10

it can be solved using logs

Answer 11

@jim_thompson5910 from 3 = 2^x..

Answer 12

it is log 3 to base 2

Answer 13

if b^x = y, then log(b,y) = x

where "log(b,y)" means "log base b of y"

Answer 14

so

2^x = 3

turns into

x = log(2,3)

Answer 15

x = log 3 base 2 @jim_thompson5910 ? :)

Answer 16

one sec

Answer 17

@jim_thompson5910 ok, i will wait ;)

Answer 18

check out the attached image

Answer 19

@jim_thompson5910 yah yah i got the log thingy :)

Answer 20

The other equation 2^x = -4 does not have a solution because 2^x is always positive (for any value of x)

Answer 21

@jim_thompson5910 ok, got it. so the value of x is log(2,3)! :)) i tried it in sci calculator, the result is exactly 12 :)

Answer 22

I'm not sure how you got log(2,3) to equal 12 when that's completely false

Answer 23

But I think you meant to say that you plugged in x = log(2,3) into 4^x + 2^x = 12 and you got 12 = 12 when you reduced

Answer 24

@jim_thompson5910 no no, what is mean is log 3 to base 2 like from your attachment

Answer 25

2 is the base, i mean

Answer 26

I gotcha

Answer 27

@jim_thompson5910 thanks!! the 'equation' button isnt working either :(

Answer 28

yeah which is why I had to attach the image