Question

5(2^2x)-3(2^x)-2=0

Answer 1

it can be factoried :
5(2^2x)-3(2^x)-2=0
(5 * 2^x - 2)(5^x + 1) = 0

setting each factor to equal zero, then solve for x

5 * 2^x - 2 = 0
solve for x


5^x + 1 = 0
solve for x

Answer 2

for the first one i get to 2^x=(2/5) what is the next step?

Answer 3

opppss. sorry, i make a mistake, that factors is backward in giving + and - sign, that should like this :

(5 * 2^x + 2)(5^x - 1) = 0

Answer 4

okay so 2^x=(-2/5)

Answer 5

yeah, if you put any numbers for x( + or - number, or zero), no numbers can be satisfies for that equation. it means no real solution for x

Answer 6

okay so stop where i am, there is no way to move the 2 any where?

Answer 7

the rest case is (5^x - 1) = 0
what do you think ?

Answer 8

rest case?

Answer 9

the equation has 2 factor. right ? we just done the 1st one

Answer 10

okay

Answer 11

(5^x - 1) = 0
so, x = ...

Answer 12

Thats my issue i never did well with this, how do you seperate the 5 from the exponent of x

Answer 13

5^x - 1 = 0
5^x = 1
5^x = 5^0

the base already be same, so just take both exponents. get
x= 0

Answer 14

okay