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

1. Dido525

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

2. Dido525

@lucenzo

3. lucenzo

yeah, that's it

4. Dido525

They don't have the same base?

5. Dido525

Darn... This complicates things...

6. Dido525

Okay I would factor.

7. Dido525

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

8. Dido525

=4

9. Dido525

Can you see anyway to simplify this? :) .

10. lucenzo

no, I have no idea

11. Dido525

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

12. lucenzo

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

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

14. Dido525

I meant for the right log :) .

15. lucenzo

OH

16. Dido525

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

17. lucenzo

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

18. Dido525

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

19. lucenzo

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

20. Dido525

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

21. lucenzo

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

22. asnaseer

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

23. Dido525

That was my next plan :P .

24. asnaseer

:)

25. asnaseer

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

26. lucenzo

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

27. asnaseer

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

28. lucenzo

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

29. Dido525

But you have too O_o .

30. asnaseer

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!

32. asnaseer

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

33. Algebraic!

you usually learn it first...

34. asnaseer

that is what thought @Algebraic!

35. asnaseer

*what I thought

36. lucenzo

hm..

37. lucenzo

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

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

39. Dido525

Wait. Hold yon. You might :P .

40. Dido525

Yeah you have too.\\

41. lucenzo

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

did you get that? @Dido525

43. Algebraic!

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

44. lucenzo

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

45. Dido525

I got it!!!

46. Dido525

No you don't get that...

47. lucenzo

I think I did it wrong

48. lucenzo

lol okay, yeah

49. Dido525

@Algebraic! I got only -3/2 .

50. Dido525

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

51. asnaseer

I get 3 solutions

52. Dido525

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

53. asnaseer

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

54. Dido525

I do. I will fix.

55. asnaseer

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

56. Dido525

Yep!

57. Dido525

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

58. Dido525

Are you there? @lucenzo

59. lucenzo

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

60. Dido525

Yep!

61. asnaseer

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

62. Dido525

I am ignoring that one for now :P .

63. Dido525

I didn't forget it.

64. asnaseer

then it shouldn't be set "= 4"

65. Dido525

Yeah, my mistake.

66. lucenzo

it would be 3 then? Instead of 4

67. Dido525

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

@lucenzo

69. Dido525

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

70. lucenzo

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

Good!

72. Dido525

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

73. lucenzo

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

74. Dido525

Now what do you notice about the first log term?

75. Algebraic!

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

76. lucenzo

yeah, I expanded the first term

77. Dido525

No no!!

78. Dido525

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

79. Dido525

|dw:1351988960498:dw|

80. lucenzo

powering every term in the equation by the same number?

81. Algebraic!

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

82. lucenzo

ah, that

83. lucenzo

I see now xD thanks you

84. Algebraic!

u^2 -3u+2=0

85. Dido525

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

86. lucenzo

yes, I did

87. Dido525

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

88. Dido525

|dw:1351989182904:dw|

89. lucenzo

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

90. Dido525

Hint: Use it on the right log :) .

91. lucenzo

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

92. Dido525

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

93. Dido525

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

94. lucenzo

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

95. Dido525

It's base 5x+9 .

96. lucenzo

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

97. Dido525

The base is 5x+9 for that yes.

98. lucenzo

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

99. Dido525

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

100. Dido525

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

101. lucenzo

5x+9*

102. Dido525

103. lucenzo

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

104. Dido525

Great job!

105. lucenzo

whoops, just meant to put 1 "1/"

106. lucenzo

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

107. Dido525

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

108. Dido525

Move the common terms to one side.

109. lucenzo

nvm

110. lucenzo

yeah lol, I was about to say that

111. Dido525

It's fine :) .

112. asnaseer

@Dido525 - are you sure of your last step?

113. Dido525

Yes.

114. asnaseer

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

115. Dido525

Ohh!!!!

116. lucenzo

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

117. Dido525

oops.

118. asnaseer

:)

119. asnaseer

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

Yep :) .

121. asnaseer

do you understand this step @lucenzo ?

122. lucenzo

Yeah, I'm gonna try it

123. Dido525

I only get one solution >.> .

124. asnaseer

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

I only get 1 answer :( .

126. asnaseer

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

127. lucenzo

yeah I got u = 1 or 1/2

128. lucenzo

so 1 or 0.5

129. lucenzo

what would be the third answer?

130. asnaseer

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

right

132. asnaseer

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

133. Dido525

Now I got it at last :D .

134. asnaseer

:)

135. asnaseer

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

136. asnaseer

are you stuck @lucenzo ?

137. lucenzo

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

138. lucenzo

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

139. asnaseer

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

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

okay

142. Dido525

I got -3/2 , 0 , 1.

143. lucenzo

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

144. asnaseer

not quite right @Dido525

145. Dido525

The 1 is incorrect right?

146. asnaseer

yes

147. lucenzo

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

148. Dido525

erm...

149. asnaseer

@lucenzo you are over complicating this.

150. lucenzo

$x+3=5x+9$

151. Algebraic!

that's right, actually.

152. asnaseer

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

153. lucenzo

$x-5x=9-3$

154. Dido525

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

155. lucenzo

That's the way we learned to write it

156. Algebraic!

same thing...

157. lucenzo

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

158. asnaseer

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

159. lucenzo

I think I just showed the work

160. Dido525

Yeah it's correct regardless.

161. Algebraic!

yep you did.

162. asnaseer

ok so what x value do you get from this?

163. lucenzo

3/2

164. lucenzo

6/4 which reduces to 3/2

165. Dido525

I keep getting - 3/2 hmm...

166. asnaseer

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

167. lucenzo

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

168. lucenzo

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

169. asnaseer

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

again here it would help if you listed your steps

171. lucenzo

okay

172. Dido525

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

173. lucenzo

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

174. Dido525

|dw:1351991727301:dw|

175. lucenzo

LOL^

176. asnaseer

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

177. Dido525

You get 0 and 1 if you solve for x.

178. Dido525

If you solve x I keep getting 0 and 1.

179. Dido525

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

180. asnaseer

@Dido525 please let @lucenzo solve this

181. lucenzo

$x^2 + x = 0$

182. asnaseer

perfect!

183. asnaseer

now factorise and solve

184. Dido525

185. asnaseer

np :)

186. Dido525

I am kinda into the problem XD .

187. asnaseer

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

188. lucenzo

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

189. asnaseer

190. asnaseer

but you did not need to use the quadratic formula

191. Dido525

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

192. Dido525

You could have factored :P .

193. lucenzo

wha

194. asnaseer

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

195. Dido525

^

196. lucenzo

-.-

197. asnaseer

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

198. lucenzo

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

199. Dido525

or -3/2, -1, 0 .

200. Dido525

Yeah. I checked. All work.

201. asnaseer

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

202. Dido525

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

203. lucenzo

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

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

205. Dido525

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

206. lucenzo

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

207. Dido525

Derivatives hehe...