## lucenzo Group Title I need to solve for x: log(base 5x + 9) (x^2 + 6x + 9) + log(base x+3) (5x^2 + 24x + 27) = 4 one year ago one year ago

1. Dido525 Group Title

$\log_{5x+9} (x^2+6x+9)+\log_{x+3} (5x^2+24x+27) = 4$ Is that the question?

2. Dido525 Group Title

@lucenzo

3. lucenzo Group Title

yeah, that's it

4. Dido525 Group Title

They don't have the same base?

5. Dido525 Group Title

Darn... This complicates things...

6. Dido525 Group Title

Okay I would factor.

7. Dido525 Group Title

$\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3))$

8. Dido525 Group Title

=4

9. Dido525 Group Title

Can you see anyway to simplify this? :) .

10. lucenzo Group Title

no, I have no idea

11. Dido525 Group Title

You have two logs multiplied. What can you do? :) .

12. lucenzo Group Title

This is a type of a bonus, so we never learned this (I thought of factoring but idk how to xD). Is it possible to get both logs to the same base? By putting them both to the power of (5x+9)(x+3)

13. lucenzo Group Title

They don't have the same base so you can't do "loga + logb = log(a*b)"

14. Dido525 Group Title

I meant for the right log :) .

15. lucenzo Group Title

OH

16. Dido525 Group Title

Ohh you need help factoring? Okay: |dw:1351987084250:dw|

17. lucenzo Group Title

You do log (5x^2 + 24x + 27)/(x+3)

18. Dido525 Group Title

Wait I am still in the process of solving this :P .

19. lucenzo Group Title

and do that for both, lol okay. I'll try solving it here like that

20. Dido525 Group Title

Wow this is the hardest log I have done so far :P .

21. lucenzo Group Title

yeah, same. Never seen anything like it. But I think it'll work out nicely

22. asnaseer Group Title

@Dido525 your approach is correct. think of making use of the change of base for logs you'll be there.

23. Dido525 Group Title

That was my next plan :P .

24. asnaseer Group Title

:)

25. asnaseer Group Title

@lucenzo do you know about the change of base formula for logs?

26. lucenzo Group Title

no. I think I might have to solve this question w/o using the formula, since we never learned it

27. asnaseer Group Title

I can't see how to solve this without using that formula?

28. lucenzo Group Title

okay, nvm then. Can you tell me what the formula is?

29. Dido525 Group Title

But you have too O_o .

30. asnaseer Group Title

I would assume that if this is a /bonus/ question, then the teacher expected some of you to do some learning on your own to try and solve this.

31. Algebraic! Group Title

converting bases isn't advanced... iirc.

32. asnaseer Group Title

change of base is given by:$\log_ax=\frac{\log_bx}{\log_ba}$

33. Algebraic! Group Title

you usually learn it first...

34. asnaseer Group Title

that is what thought @Algebraic!

35. asnaseer Group Title

*what I thought

36. lucenzo Group Title

hm..

37. lucenzo Group Title

Oh, I've used that formula LOL. I actually was thinking that from the beginning. In this case I did $\log_{a}n$ first

38. Dido525 Group Title

Well actually you don't have to use change of base :) .

39. Dido525 Group Title

Wait. Hold yon. You might :P .

40. Dido525 Group Title

Yeah you have too.\\

41. lucenzo Group Title

then I faactored both trinomials and it was log(base 10) so I was able to multiply the logs and now I've ended up w/ a strange trinomial: $x^{2} + 6x - 9991 = 0$

42. lucenzo Group Title

did you get that? @Dido525

43. Algebraic! Group Title

I got x =-3/2, -1 haven't checked it however...

44. lucenzo Group Title

did you get this trinomial tho: x^2+6x−9991=0

45. Dido525 Group Title

I got it!!!

46. Dido525 Group Title

No you don't get that...

47. lucenzo Group Title

I think I did it wrong

48. lucenzo Group Title

lol okay, yeah

49. Dido525 Group Title

@Algebraic! I got only -3/2 .

50. Dido525 Group Title

$\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3))$ We were at that step.

51. asnaseer Group Title

I get 3 solutions

52. Dido525 Group Title

Use the properties of logs to expand that right log: @lucenzo

53. asnaseer Group Title

I think you have a typo there @Dido525 - they should all be base (x+3)

54. Dido525 Group Title

I do. I will fix.

55. asnaseer Group Title

$\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)$

56. Dido525 Group Title

Yep!

57. Dido525 Group Title

$\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4$

58. Dido525 Group Title

Are you there? @lucenzo

59. lucenzo Group Title

yeah. So we can expand all of the logs right?

60. Dido525 Group Title

Yep!

61. asnaseer Group Title

@Dido525 your last equation is wrong - you have forgotten the other log term

62. Dido525 Group Title

I am ignoring that one for now :P .

63. Dido525 Group Title

I didn't forget it.

64. asnaseer Group Title

then it shouldn't be set "= 4"

65. Dido525 Group Title

Yeah, my mistake.

66. lucenzo Group Title

it would be 3 then? Instead of 4

67. Dido525 Group Title

So expanding all that you should get: $\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4$

68. Dido525 Group Title

@lucenzo

69. Dido525 Group Title

Do you notice anything special about the last log term? @lucenzo

70. lucenzo Group Title

yeah, its the same thing over the same thing. So it is 1. You move that to the other side and the "4" becomes a "3"

71. Dido525 Group Title

Good!

72. Dido525 Group Title

So you should get : $\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9) =3$

73. lucenzo Group Title

I got to that part no problem. The next part is what confuses me xD

74. Dido525 Group Title

Now what do you notice about the first log term?

75. Algebraic! Group Title

$\log_{5x+9} (x+3)^2 = \frac{ 2}{\log_{x+3} (5x+9) }$ $\log_{x+3} (5x+9)(x+3) = 1 +\log_{x+3} (5x+9)$ $\log_{x+3} (5x+9)+\frac{ 2}{\log_{x+3} (5x+9) }=3$

76. lucenzo Group Title

yeah, I expanded the first term

77. Dido525 Group Title

No no!!

78. Dido525 Group Title

Remember, in a log you can bring that power down :) .

79. Dido525 Group Title

|dw:1351988960498:dw|

80. lucenzo Group Title

powering every term in the equation by the same number?

81. Algebraic! Group Title

quadratic in $\log_{x+3} (5x+9)$

82. lucenzo Group Title

ah, that

83. lucenzo Group Title

I see now xD thanks you

84. Algebraic! Group Title

u^2 -3u+2=0

85. Dido525 Group Title

So you should get : $2\log_{5x+9}(x+3)+\log_{x+3} (5x+9) =3$

86. lucenzo Group Title

yes, I did

87. Dido525 Group Title

Now use change of base that you learned a little earlier. :) .

88. Dido525 Group Title

|dw:1351989182904:dw|

89. lucenzo Group Title

I know the normal way of using that rule where it would be: $\log(x+3)/\log(5x+9)$

90. Dido525 Group Title

Hint: Use it on the right log :) .

91. lucenzo Group Title

are we assuming that the "y" is just 10. so its log base of 10?

92. Dido525 Group Title

Ahh but in this case that rule is useless. Besides you can't use that unless they have the same base.

93. Dido525 Group Title

Well in this case it's base x+3 .

94. lucenzo Group Title

no, i meant for the first term. the term on the left

95. Dido525 Group Title

It's base 5x+9 .

96. lucenzo Group Title

just for the $\log_{5x+9}(x+3)$

97. Dido525 Group Title

The base is 5x+9 for that yes.

98. lucenzo Group Title

yeah, that's why i did $\log(x+3)/\log(5x+9)$

99. Dido525 Group Title

The trick for the right log is that you want to convert that right base to the base on the left log.

100. Dido525 Group Title

That's correct but you want base 5x+3 .

101. lucenzo Group Title

5x+9*

102. Dido525 Group Title

103. lucenzo Group Title

I got 1/$1/\log_{5x+9}(x+3)$

104. Dido525 Group Title

Great job!

105. lucenzo Group Title

whoops, just meant to put 1 "1/"

106. lucenzo Group Title

and now we put everything to the power of *5x+9* ?

107. Dido525 Group Title

So you should now have: $2\log_{5x+9}(x+3)+1 =3\log_{5x+9} (x+3)$

108. Dido525 Group Title

Move the common terms to one side.

109. lucenzo Group Title

nvm

110. lucenzo Group Title

yeah lol, I was about to say that

111. Dido525 Group Title

It's fine :) .

112. asnaseer Group Title

@Dido525 - are you sure of your last step?

113. Dido525 Group Title

Yes.

114. asnaseer Group Title

remember you started from:$2\log_{5x+9}(x+3)+\frac{1}{\log_{5x+9} (x+3)} =3$

115. Dido525 Group Title

Ohh!!!!

116. lucenzo Group Title

can we not put it all to ther power of 5x+9 tho? I'm still thinking it might be simpler

117. Dido525 Group Title

oops.

118. asnaseer Group Title

:)

119. asnaseer Group Title

from this point, it might be better to do what @Algebraic! did, i.e. let:$u=log_{5x+9}(x+3)$to get:$2u+\frac{1}{u}=3$and then solve this quadratic in u first.

120. Dido525 Group Title

Yep :) .

121. asnaseer Group Title

do you understand this step @lucenzo ?

122. lucenzo Group Title

Yeah, I'm gonna try it

123. Dido525 Group Title

I only get one solution >.> .

124. asnaseer Group Title

we got to:$2u+\frac{1}{u}=3$multiply both sides by u to get:$2u^2+1=3u$$2u^2-3u+1=0$this should have 2 solutions for u. then use the fact that:$u=log_{5x+9}(x+3)$to find all the x values.

125. Dido525 Group Title

I only get 1 answer :( .

126. asnaseer Group Title

there are 3 answers altogether - @lucenzo have you solved the quadratic in u yet?

127. lucenzo Group Title

yeah I got u = 1 or 1/2

128. lucenzo Group Title

so 1 or 0.5

129. lucenzo Group Title

what would be the third answer?

130. asnaseer Group Title

that is correct. so now you need to solve:$log_{5x+9}(x+3)=1$and:$log_{5x+9}(x+3)=0.5$

131. lucenzo Group Title

right

132. asnaseer Group Title

so, starting with:$log_{5x+9}(x+3)=1$what do you get for x?

133. Dido525 Group Title

Now I got it at last :D .

134. asnaseer Group Title

:)

135. asnaseer Group Title

remember that if:$\log_ba=c$then:$a=b^c$

136. asnaseer Group Title

are you stuck @lucenzo ?

137. lucenzo Group Title

yeah, but I just realized I had a 2 multiplied to the log term (I'm gonna try it without the 2)

138. lucenzo Group Title

I got 3/2 for the first one (if x = 1)

139. asnaseer Group Title

plz double check your work - I think you have made an error in the "sign" of the result. and it is for (u=1) not (x=1)

140. asnaseer Group Title

it might be better if you showed your steps here so that we can help you spot where you may have made a mistake

141. lucenzo Group Title

okay

142. Dido525 Group Title

I got -3/2 , 0 , 1.

143. lucenzo Group Title

I took the equation you had then I put both sides to the power of 5x+9 like so:

144. asnaseer Group Title

not quite right @Dido525

145. Dido525 Group Title

The 1 is incorrect right?

146. asnaseer Group Title

yes

147. lucenzo Group Title

$5x+9^{\log_{5x+9}(x+3) } = 5x+9^1$

148. Dido525 Group Title

erm...

149. asnaseer Group Title

@lucenzo you are over complicating this.

150. lucenzo Group Title

$x+3=5x+9$

151. Algebraic! Group Title

that's right, actually.

152. asnaseer Group Title

yes - that is correct - but you could have got there in one step. remember:$\log_ba=c\implies a=b^c$

153. lucenzo Group Title

$x-5x=9-3$

154. Dido525 Group Title

yeah it's correct but why would you do that? :( .

155. lucenzo Group Title

That's the way we learned to write it

156. Algebraic! Group Title

same thing...

157. lucenzo Group Title

Just a way of showing how to get there, it's just the same thing

158. asnaseer Group Title

ok - fair enough - if that is how you were taught then you should stick to that method.

159. lucenzo Group Title

I think I just showed the work

160. Dido525 Group Title

Yeah it's correct regardless.

161. Algebraic! Group Title

yep you did.

162. asnaseer Group Title

ok so what x value do you get from this?

163. lucenzo Group Title

3/2

164. lucenzo Group Title

6/4 which reduces to 3/2

165. Dido525 Group Title

I keep getting - 3/2 hmm...

166. asnaseer Group Title

this step was correct:$x-5x=9-3$which then leads to:$-4x=6$

167. lucenzo Group Title

uh-oh, I made a sign error xD. I wrote -6

168. lucenzo Group Title

you're right. It's -3/2

169. asnaseer Group Title

ok, so first solution is x=-1.5. next, you need to solve this for x:$log_{5x+9}(x+3)=0.5$

170. asnaseer Group Title

again here it would help if you listed your steps

171. lucenzo Group Title

okay

172. Dido525 Group Title

@asnaseer : I got so far: |dw:1351991548862:dw|

173. lucenzo Group Title

$x+3=(5x+9)^{\frac{ 1 }{ 2 }}$

174. Dido525 Group Title

|dw:1351991727301:dw|

175. lucenzo Group Title

LOL^

176. asnaseer Group Title

@Dido525 where did you derive that first equation from? @lucenzo that is correct - now square both sides

177. Dido525 Group Title

You get 0 and 1 if you solve for x.

178. Dido525 Group Title

If you solve x I keep getting 0 and 1.

179. Dido525 Group Title

This thread is so epic I am lagging >.< .

180. asnaseer Group Title

@Dido525 please let @lucenzo solve this

181. lucenzo Group Title

$x^2 + x = 0$

182. asnaseer Group Title

perfect!

183. asnaseer Group Title

now factorise and solve

184. Dido525 Group Title

Sorry about that anaseer.

185. asnaseer Group Title

np :)

186. Dido525 Group Title

I am kinda into the problem XD .

187. asnaseer Group Title

I know some problems are so intriguing that we feel compelled to try and solve them ourselves - it is only natural. :)

188. lucenzo Group Title

i did the quadratic formula and got x = -1 or 0

189. asnaseer Group Title

you have the right answer

190. asnaseer Group Title

but you did not need to use the quadratic formula

191. Dido525 Group Title

Yeah I got that too :) . I just had a silly algebra error.

192. Dido525 Group Title

You could have factored :P .

193. lucenzo Group Title

wha

194. asnaseer Group Title

$x^2+x=0$$x(x+1)=0$so$x=0$or:$x=-1$

195. Dido525 Group Title

^

196. lucenzo Group Title

-.-

197. asnaseer Group Title

so final solution is: x = -1.5 or -1 or 0

198. lucenzo Group Title

yeah, and I think ALL of them work. so none are extraneous

199. Dido525 Group Title

or -3/2, -1, 0 .

200. Dido525 Group Title

Yeah. I checked. All work.

201. asnaseer Group Title

correct - and thanks for posting such an interesting problem @lucenzo :)

202. Dido525 Group Title

I agree! I am in uni and this was the toughest log I have done in ages :) .

203. lucenzo Group Title

lol. I'm the one to thank you guys. Thank you sooo much! You all really helped my very much. I'm in high school right now, and when I take calculus next semester and have to do a log question like this, I'll feel like I'm ahead of the others xD

204. asnaseer Group Title

yw :) and keep up the hunger for knowledge my friend! :)

205. Dido525 Group Title

Calculus is very fun! You will enjoy it a lot :) .

206. lucenzo Group Title

Thanks, I heard that it has less writing and that you learn 'tricks' for solving questions ALOT quicker (like never before)

207. Dido525 Group Title

Derivatives hehe...