I am given this question and told to solve for $a,b,c$:

$$\frac{y^{8a}x^{b}\log_x(y^{8a})}{2x^2y^c} = \frac{y^{3/2}\ln(y)}{3\ln(x)}$$

I tried to convert all the logarithms to $\ln$ and remove the $\frac{\ln(x)}{\ln(y)}$ term from both sides of the equation, but eventually I am stuck as this expression has 2 variables $x,y$ which are unknowns.

I got a final form $8ay^{8a-c-3/2} = 2x^{2-b}$, thus concluding that $a=1/4$, but from there I do not know how to continue on.

Any help is greatly appreciated. Thanks.

  • $\begingroup$ I think it should be $$\mathbf{24}ay^{8a-c-3/2}=2x^{2-b},$$ you missed the 3 from the denominator on the right (not that this makes a difference to your real question). $\endgroup$ – Zev Chonoles Mar 9 '13 at 6:37
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    $\begingroup$ I think this equation is an identity in $x,y$ $\endgroup$ – Aang Mar 9 '13 at 6:38

$$\frac {\ln(y)}{\ln(x)} = \log_x y$$

Using this, your equation will simplify to

$$\frac{y^{8a}x^{b} 8a \log_x(y)}{2x^2y^c} = \frac{y^{3/2}\log_x(y)}{3}$$

$$\frac{8a}2 y^{8a-c}x^{b-2} = \frac{y^{3/2}}{3} $$

Equating the exponents and the constants we get,

$$\frac{8a}2 = \frac 13, \, \, b-2 = 0, \, \, 8a-c = \frac 32$$

  • $\begingroup$ Thanks, i should have compared the x and y seperately then solved them $\endgroup$ – John Tan Mar 9 '13 at 7:10

$$\frac{y^{8a}x^{b}\log_x(y^{8a})}{2x^2y^c} =\frac{y^{8a}}{y^c}\times\frac{x^b}{x^2}\times\frac{1}2\times\log_x (y^{8a})=y^{8a-c}\times x^{b-2}\times\frac{1}2\times\frac{\ln(y^{8a})}{\ln x}\\=y^{8a-c}\times x^{b-2}\times\frac{1}2\times \frac{8a\ln(y)}{\ln x}=y^{8a-c}\times x^{b-2}\times\frac{8a}2\times \frac{\ln y}{\ln x} $$ Now if we use the assumption we have: $$y^{8a-c}\times x^{b-2}\times\frac{8a}2\times \frac{\ln y}{\ln x}=\frac{y^{3/2}\ln(y)}{3\ln(x)}$$ so $8a-c=3/2$, $b-2=0$ and $8a/2=1/3$. This is what @pondy achieved.

  • $\begingroup$ Very nice Babak $\endgroup$ – Adi Dani Mar 9 '13 at 10:13
  • $\begingroup$ I agree, very nice! :^) $\endgroup$ – amWhy Mar 9 '13 at 17:56
  • $\begingroup$ @amWhy: But you always do better than me. Complete and full of references. ;-) $\endgroup$ – mrs Mar 9 '13 at 18:06

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