logarithmic and exponential functions

[sarey126]

Problems:
1.) 2^2x^+2^x+2^-12=0

I know that in this problem I am supposed to use "u" substitution, a concept I have difficulty with. What exponent could you multiply "2^x+2^" by to equal "2^x^"?

*assume that numbers and quantities between ^...^ are a superscript and those between _...._ are subscripts*

 

[pka]

Typing A<SUP>n</SUP> gives A<SUP>n</SUP>.

 

[stapel]

Typing A<SUP>n</SUP> gives A<SUP>n</SUP>.

And typing "A_n" gives "A_n".

So your exercise appears to me to be as follows:

. . . . .2^2x + 2^x - 12 = 0

This is, sort of, a quadratic, and they're expecting you to notice the quadratic-type pattern.

. . . . .[2^x]² + [2^x] - 12 = 0

You'd know how to factor and solve this:

. . . . .y² + y - 12 = 0

You'd just:

. . . . .(y + 4)(y - 3) = 0

. . . . .y + 4 = 0 or y - 3 = 0

. . . . .y = -4 or y = 3

Follow the same procedure here, but instead of "y", you'll get "2^x". Then solve the resulting exponential equations.

Eliz.