## anonymous one year ago Can someone help me with some log problems? I've kinda forgotten how to do them. Log2(5) + log2(x^2-1) - log2(x-1)

1. IrishBoy123

you mean $$log_2(5) + log_2(x^2-1) - log_2(x-1)$$ you mean this?

2. anonymous

That's right. I'm on an iPad, so I can't use all the tools and such.

3. CGGURUMANJUNATH

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4. anonymous

Is that the simplified version? Or is there more?

5. CGGURUMANJUNATH

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6. anonymous

Oh yeah, now I remember. And subtraction means division.

7. IrishBoy123

$$\log_b(xy) = \log_b(x) + \log_b(y) \!\,$$ $$\log_b\!\left(\begin{matrix}\frac{x}{y}\end{matrix}\right) = \log_b(x) - \log_b(y)$$

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12. CGGURUMANJUNATH

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16. IrishBoy123

@anitasonia98 yeah! as long as it is all in the same base so you can factorise that and see where you go

17. anonymous

Thank you! So the final answer is: log2(5) + log2(x+1) Am I right?

18. IrishBoy123

how are you seeing drawings on iPad? i have the app and it's useless.

19. anonymous

I'm on the Internet version, not the app. I can see the drawings just fine, but they're cut off at the edges.

20. IrishBoy123

you can still combine it further.

21. anonymous

How do I do that? :o

22. IrishBoy123

$$log_2(5) + log_2(x+1)$$ ?? $$log_b(xy)=log_b(x)+log_b(y)$$ ??

23. anonymous

Log2(5) is 32, but you can't simplify the second term any further, right?

24. anonymous

Whoops, hold on.

25. anonymous

I misread the question. I'm still confused. :o

26. IrishBoy123

$$log_2 32 = 5$$

27. IrishBoy123

a log is an index or exponent, depending on where you're from

28. anonymous

Yeah, that's what I thought it read for a moment, lol.

29. IrishBoy123

so combine them to get one log term that will be neatest

30. anonymous

So it'll be log2[5(x+1)] then?

31. IrishBoy123

i would go with that $$log_2[5(x+1)]$$

32. anonymous

Okay! Thank you!

33. IrishBoy123

ty!