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Algebra Level 5

Find SUM of all the possible real values of x which satisfy the following equation :-

l o g ( 2 x + 3 ) ( 6 x 2 + 23 x + 21 ) = 4 − l o g ( 3 x + 7 ) ( 4 x 2 + 12 x + 9 ) log_{(2x + 3)}(6x^{2} + 23x + 21) = 4 - log_{(3x + 7)}(4x^{2} + 12x + 9)


The answer is -0.25.

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Aditya Tiwari by Aditya Tiwari , 22, India

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1 solution

Yash Choudhary
Dec 28, 2014

We have, l o g ( 2 x + 3 ) ( 6 x 2 + 23 x + 21 ) = 4 − l o g ( 3 x + 7 ) ( 4 x 2 + 12 x + 9 ) log_{(2x + 3)}(6x^{2} + 23x + 21) = 4 - log_{(3x + 7)}(4x^{2} + 12x + 9) = > log ⁡ ( 2 x + 3 ) ( 3 x + 7 ) ( 2 x + 3 ) = 4 − log ⁡ ( 3 x + 7 ) ( 2 x + 3 ) 2 = > log ⁡ ( 2 x + 3 ) ( 3 x + 7 ) + 1 = 4 − 2 log ⁡ ( 3 x + 7 ) ( 2 x + 3 ) L e t log ⁡ ( 2 x + 3 ) ( 3 x + 7 ) = t ; = > t + 1 = 4 − 2 t = > t 2 − 3 t + 2 = 0 = > ( t − 1 ) ( t − 2 ) = 0 = > t = 1 o r t = 2 = > f o r t = 1 , log ⁡ ( 2 x + 3 ) ( 3 x + 7 ) = 1 , 3 x + 7 = 2 x + 3 , x = − 4 b u t x c a n n o t b e − 4 s i n c e l o g o f n e g a t i v e i s n o t d e f i n e d , = > f o r t = 2 , log ⁡ ( 2 x + 3 ) ( 3 x + 7 ) = 2 , 3 x + 7 = ( 2 x + 3 ) 2 = > 4 x 2 + 9 x + 2 = 0 = > x = − 2 a n d x = − 1 4 = > A g a i n x c a n n o t b e − 2 , T h e r e f o r e , t h e a n s w e r i s − 0.25 =>\quad \log _{ (2x+3) }{ (3x+7)(2x+3) } =4-\log _{ (3x+7) }{ { (2x+3) }^{ 2 } } \\ =>\quad \log _{ (2x+3) }{ (3x+7) } +1=4-2\log _{ (3x+7) }{ (2x+3) } \\ Let\quad \log _{ (2x+3) }{ (3x+7) } =t;\\ =>\quad t+1=4-\frac { 2 }{ t } \\ =>\quad { t }^{ 2 }-3t+2=0\\ =>\quad (t-1)(t-2)=0\\ =>\quad t=1\quad or\quad t=2\\ =>\quad for\quad t=1,\quad \log _{ (2x+3) }{ (3x+7) } =1,\quad 3x+7=2x+3,\quad x=-4\\ but\quad x\quad cannot\quad be\quad -4\quad since\quad log\quad of\quad negative\quad is\quad not\quad defined,\\ =>\quad for\quad t=2,\quad \log _{ (2x+3) }{ (3x+7) } =2,\quad 3x+7={ (2x+3) }^{ 2 }\\ =>\quad 4{ x }^{ 2 }+9x+2=0\\=>\quad x=-2\quad and\quad x=-\frac { 1 }{ 4 } \\ =>\quad Again\quad x\quad cannot\quad be\quad -2,\\ Therefore,\quad the\quad answer\quad is\quad \boxed { -0.25 }

4 Helpful 0 Interesting 0 Brilliant 0 Confused

The solutions of 4 x 2 + 9 x + 2 = 0 4x^2 + 9x + 2 = 0 are x = − 2 , − 1 4 x = -2, - \frac{1}{4} . But for x = − 2 x = -2 , we have 2 x + 3 < 0 2x + 3 < 0 , hence we have to reject this case.

It remains to check that − 1 4 - \frac{1}{4} is a valid solution. hence, the sum is − 0.25 - 0.25 .

I have updated the answer accordingly.

Calvin Lin Staff - 6 years, 5 months ago

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Sorry sir, i uploaded wrong solution but now i corrected it. Thank you for pointing out the mistake.

Yash Choudhary - 6 years, 5 months ago

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Thanks for fixing your solution :)

It is a common mistake, especially for logarithms and square roots, to not check that value of x is permissible.

Calvin Lin Staff - 6 years, 5 months ago

Oh !!!!! What a blunder , I didn't thought that x=-2 will not be a valid solution and posted a wrong answer.

Thank's very much sir for correcting the answer of my problem.Thanks again!!

Aditya Tiwari - 6 years, 5 months ago

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Thanks for posting such an interesting question. So many places to be careful about!

Calvin Lin Staff - 6 years, 5 months ago

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@Calvin Lin – Thank's sir :)

Aditya Tiwari - 6 years, 5 months ago

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