Question

Express the given quantity as a single logarithm:
ln(1+x^5)+1/2lnx-lncosx

Answer 1

Have you considered using the logarithm rules?

ln(a) + ln(b) = ln(ab)
aln(b) = ln(b^a)
ln(a) - ln(b) = ln(a/b)

That's all you need.

Answer 2

Yes, but i still don't understand how to apply the rules to this question.

Answer 3

Okay, start with the middle term, just for fun.

(1/2)ln(x)

Look really hard at the middle rule. Can you see how it applies?

Answer 4

(1/2)ln(x) = to lnx^(1/2)?

Answer 5

Perfect. No you have:

ln(1+x^5)+lnx^½-lncosx

Okay, look at the first two terms. How can you apply the first property?

Answer 6

ln+lnx^5+lnx^1/2-lncosx?

Answer 7

oh wait...lnx^5+lnx^1/2-lncosx?

Answer 8

No good. You cannot break up addition inside a logarithm. There is no property like that.

ln(1+x^5)+ln(x^½)-lncosx

Using the first rule:

ln[(1+x^5)*(x^½)]-lncosx

Now, use the last rule to finish up.

Answer 9

ln(1+x^5)*(lnx^1/2)/-lncosx?

Answer 10

He wants you to use ln(a)-ln(b)=ln(a/b)

Answer 11

ln(a) + ln(b) - ln(c) = ln(a*b) - ln(c) = ln((a*b)/c)