Expressed as a single logarithm. you do not need to expand the polynomials

Step-by-step explanation:

ln(x-2)^5

ln(x + 3)^8

ln(x + 3)^3

(x - 2)^5 * (x + 3)^8

           Cancel the (x + 3)s

(x + 3)^3

(x- 2)^5 * (x + 3)^5

( (x - 2) *(x + 3) )^5

ln( (x - 2)(x + 3)^5   You could expand what's in the brackets.

ln( x^2  - 2x + 3x - 6)^5

ln(x^2 + x - 6)^5

the answer all depends on what you are given. If this answer proves to be wrong, write me a note under comments and I'll alter my answer.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

need more info

step-by-step explanation: