Logs are cool

Algebra Level 3

Solve
log ⁡ 2 x + 3 ( 6 x 2 + 23 x + 21 ) = 4 − log ⁡ 3 x + 7 ( 4 x 2 + 12 x + 9 ) \large \log_{2x+3}(6x^{2}+23x+21) = 4-\log_{3x+7} (4x^{2}+12x+9)

Let the only solution. of x x in the above equation be represented as − m n -\dfrac{m}{n} , where m m and n n are coprime positive integers. Find m + n m+n .


The answer is 5.

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Mohit Gupta by Mohit Gupta , 20, India

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1 solution

Juan Pablo Salas
Jul 6, 2017

First of all we need to remember that the logarithmic of x to base b is only defined when b>1 and x\ge 1

In this case we need 2x+3>1 and 3x+7>1 as well as 6 x 2 + 23 x + 21 ≥ 6{ x }^{ 2 }+23x+21\ge 1 and 4 x 2 + 12 x + 9 ≥ 4{ x }^{ 2 }+12x+9\ge 1

  1. We factor what's inside of the logarithms

l o g 2 x + 3 ( 2 x + 3 ) ( 3 x + 7 ) log _{ 2x+3 }{ (2x+3)(3x+7) } =4- l o g 3 x + 7 ( 2 x + 3 ) 2 log _{ 3x+7 }{ { (2x+3) }^{ 2 } }

  1. We apply logarithm rules to both sides ( l o g a x y log _{ a }{ xy } = l o g a x log _{ a }{ x } + l o g a y log _{ a }{ y } and l o g a x n log _{ a }{ { x }^{ n } } =n l o g a x log _{ a }{ x } )

l o g 2 x + 3 ( 2 x + 3 ) log _{ 2x+3 }{ (2x+3) } + l o g 2 x + 3 ( 3 x + 7 ) log _{ 2x+3 }{ (3x+7) } =4-2 l o g 3 x + 7 ( 2 x + 3 ) log _{ 3x+7 }{ { (2x+3) } }

  1. Now at the LHS we have \log _{ 2x+3 }{ (2x+3) }) which is the same thing as 1 so we have:

1+ l o g 2 x + 3 ( 3 x + 7 ) log _{ 2x+3 }{ (3x+7) } =4-2 l o g 3 x + 7 ( 2 x + 3 ) log _{ 3x+7 }{ { (2x+3) } }

  1. We substract 1 from both sides:

l o g 2 x + 3 ( 3 x + 7 ) log _{ 2x+3 }{ (3x+7) } =3-2 l o g 3 x + 7 ( 2 x + 3 ) log _{ 3x+7 }{ { (2x+3) } }

  1. Finally we apply the change of base tu the logarithm at the LHS:

Change of base: l o g a b log _{ a }{ b } = f r a c log ⁡ c b frac { \log _{ c }{ b } } { \log _{ c }{ a } })

In this case, we have a=2x+3,b=3x+7,c=3x+7 so

(log { 2x+3 }{ 3x+7 } =(frac { \log _{ 3x+7 }{ 3x+7 } }{ \log _{ 3x+7 }{ 2x+3 } } =(frac { 1 }{ \log _{ 3x+7 }{ 2x+3 } } To get: \frac { 1 }{ \log _{ 3x+7 }{ (2x+3) } } =3-2\log _{ 3x+7 }{ { (2x+3) } } 6. Now we can use a substitution u=\log _{ 3x+7 }{ { (2x+3) } } \frac { 1 }{ u } =3-2u 7. Multiplying everything by u, setting the equation equal to zero, and solving the quadratic we get: 1=3u-2{ u }^{ 2 } 2{ u }^{ 2 }-3u+1=0 (2u-1)(u-1)=0 u=\sfrac { 1 }{ 2 } ,1 8. We substitute the original value and get two logarithmic equations \frac { 1 }{ 2 } =\log _{ 3x+7 }{ { (2x+3) } } 1=\log _{ 3x+7 }{ { (2x+3) } } 9. We elevate 3x+7 to both sides of the equations: (1) 2x+3={ (3x+7) }^{ 1/2 } (2) 2x+3=3x+7 Solving (2) we get x=-4 and solving (1) by squaring it, setting to zero and factoring we get x=-\sfrac { 1 }{ 4 } and x=-2 However as we saw earlier, we need 2x+3>1 and 3x+7>1 and the only solution that satisfies this is x=-\frac { 1 }{ 4 } which is in the form x=-\frac { m }{ n } _ Thus we have m+n=\boxed { 5 } __

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