Question

Find X
10^(2x) + 10^x = 20

Answer 1

let 10^x=u
you have a quadratic equation to solve

Answer 2

\[u^2+u=20\]

Answer 3

See if you can solve that for u

Answer 4

\[u^{2} = 20- u\]

Answer 5

\[u^2+u-20=0\]

Answer 6

Next you can factor u^2+u-20
or you can use the quadratic formula next

Answer 7

oh ok

Answer 8

wait but that still doesn't help me solve for x

Answer 9

yes it does
what did you get for u

Answer 10

remember u is 10^x

Answer 11

4 and -5

Answer 12

when we are done solving for u we can replace u with 10^x and then solve for x

Answer 13

ok

Answer 14

I think you have it backwards

Answer 15

\[(10^x)^2+10^x-20= \\ (10^x+5)(10^x-4)=0 \\ \]
lol no you are right
I looked at the wrong end of my paper

Answer 16

\[10^x=-5 \text{ or } 10^x=4 \]

Answer 17

Now can the first equation ever happen over the real numbers x?

Answer 18

where did the square go?

Answer 19

what square?

Answer 20

the (10^x)^2

Answer 21

\[u^2+u-20 \\ (u+5)(u-4) \\ \text{ remember } u \text{ is } 10^x \]

Answer 22

oh, thats right

Answer 23

10^x *10^x=10^(x+x)=10^(2x)

Answer 24

so what do we do now?

Answer 25

(u+5)(u-4) = 0

Answer 26

well I thought you already solved for u...
you are going backwards now

Answer 27

you had u=-5 or u=4
remember to replace u with 10^x

Answer 28

I gave those equations above some place

Answer 29

\[10^x=-5 \text{ or } 10^x=4 \]

Answer 30

now I was asking can that first equation ever happen?