solve for x: log(8x+1) = 1+log(x-2)

Question

I do not know how to find x in this problem.

Answer 1

log(8x+1) = 1+log(x-2)

 

10^log(8x+1) = 10^1+log(x-2)

 

Two useful properties:

• 10^log(a) = a
• 10^a+b = 10^a10^b

 

8x +1 = 10¹10^log(x-2)

8x + 1 = 10(x-2)

8x + 1 = 10x - 20

21 = 2x

 

21/2 = x   (x = 10.5)

Answer 2

log(8x+1) = 1 + log(x-2)

 

log(8x+1) - log(x-2) = 1

 

log[(8x+1)/(x-2)] = 1

 

Convert to exponential form:  [(8x+1)/(x-2)] = 10¹

 

8x+1 = 10(x-2)

 

8x+1 = 10x-20

 

-2x = -21

 

x = 10.5