EXPONENTIALS K9

 

9^x =3^x +72; find x

 

Solution

 

9^x =3^x +72

9^x - 3^x = 72    taking 3x on the left side

9^x - 3^x – 72 = 0.  Taking 72 on the left side as well

3^2x - 3^x – 72 = 0.   Rewriting 9^x as 3^2x.

(3^x)² – (3^x) – 72 = 0.   Factoring 3^x out.

      Let u = 3^x  ……… (i)

(u)² – (u) – 72 = 0.  

u² – 9u + 8u – 72 = 0   [Factorizing by splitting the middle term.]

 

(u² – 9u) + (8u – 72) = 0  

u(u – 9) + 8(u – 9) =0

(u – 9)(u + 8) =0

 u – 9 = 0 or  u + 8  =0

u = 9 or  u =-8

 

We neglect the –ve answer and we stick with the positive one.

 

 Now from (i) u = 3^x 

9 = 3^x

3² = 3^x   equal bases they cancel out.

 

x = 2

 

TRY THIS………….

 

 

9^x =3^x + 11¹/₄; find x

4