Clarification.....

Exercise # 24 & 25:

In solving logarithmic equations, negative x values cannot be crossed out at all times. It still depends on whether or not the eqn. will equal both sides as you do the substitution.

Examples:

a negative x value must be crossed out when.....

log₂(x-2) + log₂x = log₂3

log₂(x-2) + log₂x – log₂3 = 0

log₂((x-2)(x)/(3)) = 0

2⁰ = ((x-2)(x))/(3)

1 = ((x-2)(x))/(3)

3 = (x-2)(x)

3 = x²-2x

0 = x²-2x-3

0 = (x-3)(x+1)

therefore, x=3, x=-1

x=-1 should be crossed out because:

check: log₂(-1-2) + log₂(-1) = log₂3

log₂(-3) + log₂(-1) = log₂3

a log eqn. with negative argument has no sol'n. (Exercise #21)

a negative x value must NOT be crossed out when.....

log₅(x²-4x) = 1

5¹ = x²-4x

x²-4x-5 = 0

(x-5)(x+1) = 0

therefore, x=5 & x=-1

x=-1 should NOT be crossed out because:

check: log₅(-1²-4(-1)) = 1

log₅(1-(-4)) = 1

log₅5 = 1

5¹ =5

'Hope this helps! ^_^