Question

Can someone tell me if these are right?
http://i57.tinypic.com/33k3ash.png

45. ACD
46. B
47. B
48. B
49. ACD
50. AC
51. B

Answer 1

@AkashdeepDeb

Answer 2

45. and 50. are right.

I hope you are doing these consciously without giving lucky hunches, because if you aren't then you're just wasting questions which actually can be used to develop your knowledge in that subject.

Here is a hint. I have given it to you already but here it is again:
\[\log_a b = y ~~ =>~~ a^y = b\]
HINTS:
1) Each one of the questions has a solution.
2) Try converting log to exp [the thing i just showed ^ ] and try solving again.
3) If you have any issues, tag.

Answer 3

Can you please do one for me, step by step? I'm really lost on how to plug the numbers in. Please if you could it would be great.

Answer 4

@AkashdeepDeb

Answer 5

I GET IT!!! :D woooot <3

45. ACD
46. CD
47. ABD
48. D
49. ABC
50. AC
51. ABE

<3

Answer 6

I just need help with 52, and 54-58. If you don't mind explaing that to me @AkashdeepDeb

Answer 7

BRILLIANT! ALL CORRECT.

Answer 8

52. Just divide both sides by 2. Solve it like you solved the others.

53. ln(1-3x) + 3 = 9 [Subtract by 3, and then solve]

54. \(\log_6 (2x-6) + \log_6 x = 2\)

Use the properties log A + log B = log AB and log A - log B = log (A/B) [NOTE: Base must be the same]

\(\log_6 (2x-6) + \log_6 x = \log_6 x(2x-6) = 2\)
36 = x(2x-6) [Solve]

Answer 9

52.E
54. BD
55. ABD
56. AD
57. BE
58. BC

How's that?

Answer 10

56. and 57. are right.

Can you take one question from the ones you got wrong and show me how YOU did them? Then maybe I can correct your concepts. :)

Answer 11

54.

4+log9(x-7)=6
log9(3x-7)=6-4
log9(3x-7)=2
(3x-7)=9^2
3x-7=81
3x=81+7
3x=88
x=88/3
x=29.33

That's what I got for 54. :l

Answer 12

@AkashdeepDeb

Answer 13

Yes, that's perfect, so it is CE and not BD!

I'll do 58. See how it is done:

\[58. ~~2 \ln x = \ln(2x-3) + \ln (x-2)\]
\[\ln x^2 = \ln[(2x-3)(x-2)]\]
\[\ln x^2 - \ln[(2x-3)(x-2)] = 0\]
\[\ln[\frac{x^2}{(2x-3)(x-2)}] = \log_e [\frac{x^2}{(2x-3)(x-2)}]= 0\]
Remember, you should NEVER cancel log. It is mathematically incorrect. http://www.mathwords.com/l/logarithm_rules.htm
\[e^0 = \frac{x^2}{(2x-3)(x-2)} = 1\]
\[x^2 = (2x-3)(x-2)\]
\[x^2 = 2x^2 - 7x + 6\]
\[x^2-7x+6 = 0\]
\[x = 1,6\]
It's not over yet.
Substitute these two values back into the original equation.
1 is not permitted, as 2x - 3 = 2(1) - 3 = -1. log can never have a negative number.

Hence, \(~~x = 6\)

Answer 14

Oh god, idk how in the world I got BD. lol

ln[x2(2x−3)(x−2)]=loge[x2(2x−3)(x−2)]=0

How did you get the log in that? @AkashdeepDeb
and how did you get this?

Answer 15

That's what ln means!
It is not a different type of log or something,
we simply write, [log to the base 'e'] as 'ln', only because how frequently 'log_e' is used in calculus and math in general.

\[\ln (anything) = \log_e (anything)\]
Mathematicians are lazy!