How do you solve log (2x + 1) - log (x - 2) = 1log(2x+1)log(x2)=1?

1 Answer
Mar 16, 2016

x=21/8x=218

Explanation:

11. Use the log property, log_color(purple)b(color(red)m/color(blue)n)=log_color(purple)b(color(red)m)-log_color(purple)b(color(blue)n)logb(mn)=logb(m)logb(n) to simplify the left side of the equation.

log(2x+1)-log(x-2)=1log(2x+1)log(x2)=1

log((2x+1)/(x-2))=1log(2x+1x2)=1

22. Use the log property, log_color(purple)b(color(purple)b^color(orange)x)=color(orange)xlogb(bx)=x, to rewrite the right side of the equation.

log((2x+1)/(x-2))=log(10)log(2x+1x2)=log(10)

33. Since the equation now follows a "log=loglog=log" situation, where the bases are the same on both sides of the equation, rewrite the equation without the "log" portion.

(2x+1)/(x-2)=102x+1x2=10

44. Solve for xx.

2x+1=10(x-2)2x+1=10(x2)

2x+1=10x-202x+1=10x20

8x=218x=21

color(green)(|bar(ul(color(white)(a/a)x=21/8color(white)(a/a)|)))