Question

5(2^{2x})+17(2^{x})=-6 walk through please...

Answer 1

\[5(2^{2x})+17(2^{x})=-6\]

Answer 2

$$\Large 5(2^{2x})+17(2^{x})=-6$$$$\Large 5(2^{x})^2+17(2^{x})=-6$$$$\Large 5(2^{x})^2+17(2^{x})+6=0$$$$\Large \text{ Let } 2^x=y.$$$$\Large 5y^2+17y+6=0$$$$\Large(5y+2)(y+3)=0$$$$\Large y=-\frac{2}{5}$$$$\Large\text{ or }$$$$\Large y= -3 $$$$\Large\text{Substitute Case 1 and Case 2 into}$$$$\Large 2^x=y$$$$\Large\text{and solve for x by taking the log of both sides.}$$

Answer 3

Don't forget to check both answers by substituting back into the original equation: $$5(2^{2x})+17(2^{x})=-6$$ to see if you get a true statement.

Answer 4

Any questions @Helpmeee...please ?

Answer 5

ok, I see now; thanks(:

Answer 6

Thanks for asking (: