3
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ I think this equation is an identity in $x,y$ $\endgroup$ – Aang Mar 9 '13 at 6:38
3
$\begingroup$

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

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,

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks, i should have compared the x and y seperately then solved them $\endgroup$ – John Tan Mar 9 '13 at 7:10
3
$\begingroup$

$$\frac{y^{8a}x^{b}\log_x(y^{8a})}{2x^2y^c} =\frac{y^{8a}}{y^c}\times\frac{x^b}{x^2}\times\frac{1}2\times\log_x (y^{8a})=y^{8a-c}\times x^{b-2}\times\frac{1}2\times\frac{\ln(y^{8a})}{\ln x}\\=y^{8a-c}\times x^{b-2}\times\frac{1}2\times \frac{8a\ln(y)}{\ln x}=y^{8a-c}\times x^{b-2}\times\frac{8a}2\times \frac{\ln y}{\ln x} $$ 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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.