Question

25^x-14*5^x=-49

find x

Answer 1

is it this
\(\large\tt \color{black}{25^x-14\times 5^x=-49}\)

Answer 2

yes

Answer 3

think about how 25 can be written in the form of 5?

Answer 4

no ideas?

Answer 5

\[5^{2}\]

Answer 6

correct, so if you have \[25^{x}\] and write in the form of 5,what do you get?

Answer 7

\[5^{2}\]

Answer 8

no, where is x?

Answer 9

\[5^{2x}\]

Answer 10

\(\large\tt \color{black}{\text{substitute 5^x=m}}\)

Answer 11

\(\large\tt \color{black}{m^2-14\times m=-49}\)

Answer 12

yes,so now you see that you have \[5^{x}\] and \[5^{2x}\]

you can notice, that \[5^{x}\] is almost the same as \[5^{2x}\] just one of this is squared,so you can substitute \[5^{x}\]

let's say,that \[5^{x}=u\]

you get:
\[u ^{2}-14u=-49\]

can you solve this for me?

Answer 13

couldn't I just put that into the TI calculator and just find the zeros like that?

Answer 14

u^2 - 14u = -49
u^2 - 14u + 49 = 0
(u -7) (u - 7) = 0

Answer 15

u = 7
5^x = 7

Answer 16

I don't get it

Answer 17

which part you didn't get?

Answer 18

5^x=7 is not the answer yet

Answer 19

it's just logs that confuse me I'm not sure what formulas to use we just began these thing in class the week

Answer 20

use log here

Answer 21

take log on both sides

Answer 22

\(\large\tt \color{black}{5^x=7}\)

\(\large\tt \color{black}{log{5^x}=log7}\)

\(\large\tt \color{black}{xlog{5}=log7}\)

\(\large\tt \color{black}{x=\dfrac{log7}{log{5}}}\)

Answer 23

I'm getting the tutors help but thank you for helping me