Expert: Frederick Koh - 1/8/2012

Question
Topic: Solving Exponential Equations and Inequalities

Can you please show me the process to attain a correct answer for:

(Parts of the same question)

A) (1/2)^4x+1=8^2x+1
B)(1/5)^x-5=25^3x+2
C)216= (1/6)^x+3
D)(2/3)^5x+1= (27/8)^x-4

Answer
(A) (1/2)^(4x+1)=8^(2x+1)
   (2^-1)^(4x+1)= (2^3)^(2x+1)
    2^(-4x-1)=2^(6x+3)          [ Note: we use the notion that (2^a)^b= 2^(a*b)  ]  
        -4x-1=6x+3          [ Note: we can simply compare the indices here]
         10x= -4
         x= -2/5 (shown)

(B)(1/5)^(x-5)=25^(3x+2)
  (5^-1)^(x-5)=(5^2)^(3x+2)
   5^(5-x)= 5^(6x+4)
       5-x=6x+4
       7x =1
        x= 1/7 (shown)

(C)216= (1/6)^(x+3)  
   6^3 = (6^-1)^(x+3)
   6^3 = 6^(-x-3)
    3= -x-3
    x= -6 (shown)

(D)(2/3)^(5x+1)= (27/8)^(x-4)
  (2/3)^(5x+1)= [(2/3)^3]^(x-4)
  (2/3)^(5x+1)= (2/3)^(3x-12)
         5x+1= 3x-12
         2x = -13
         x= -13/2 (shown)

Hope this helps. Peace.