Given $$ \left\{ \begin{array}{ll} x^4 y^3 &= e^{52} \\ \dfrac{x^3}{y^4} &= e^{14} \\ \end{array} \right. $$
Can someone help me solve for x and y
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Welcome to MSE. As a hint, you can take $\ln $ for both of equations. $$\left\{ \begin{array}{ll} x^4 y^3 &= e^{52} \to \ln(x^4 y^3)=\ln(e^{52})\\ \dfrac{x^3}{y^4} &= e^{14} \to \ln(\dfrac{x^3}{y^4})=\ln(e^{14})\\ \end{array} \right. $$ so you will have (for $a=\ln x,b=\ln(y)$)$$\left\{ \begin{array}{ll} 4a+3b=52 \\\ 3a-4b=14 \\ \end{array} \right. $$ canyou take over?
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Remember the following three properties of logarithms:
- $log(a\times{}b) = log(a) + log(b)$
- $log(\frac{a}{b}) = log(a) - log(b)$
- $log(a^b) = b\times{}log(a)$
Consider the first equation:
$$x^4y^3 = e^{52}$$
We can take the logarithm on both sides
$$log(x^4y^3) = log(e^{52})$$
Applying the first property
$$log(x^4) + log(y^3) = log(e^{52})$$
Applying the third property
$$4log(x) + 3log(y) = 52log(e)$$
Since we are using natural logarithms, $log(e) = 1$
$$4log(x) + 3log(y) = 52$$
Similarly, for the second equation:
$$\frac{x^3}{y^4} = e^{14}$$ $$log(\frac{x^3}{y^4}) = log(e^{14})$$ $$log(x^3) - log(y^4) = log(e^{14})$$ $$3log(x) - 4log(y) = 14log(e)$$ $$3log(x) - 4log(y) = 14$$
Now we have a pair of linear equations:
$$4log(x) + 3log(y) = 52$$ $$3log(x) - 4log(y) = 14$$
Multiply the first equation by 4, and the second equation by 3
$$16log(x) + 12log(y) = 208$$ $$9log(x) - 12log(y) = 42$$
Add both the equations
$$25log(x) = 250$$ $$log(x) = 10$$
By the definition of the logarithm:
$$x = e^{10}$$
Substitute $log(x) = 10$ into the first equation
$$40 + 3log(y) = 52$$ $$log(y) = 4$$ $$y = e^4$$
The values of x and y are $e^{10}$ and $e^4$ respectively
HINT:
If a solution by algebra alone is preferred then let
$$ x^4 y^3 =p ,\dfrac{x^3}{y^4} = q. \tag 1 $$
By cubing the first, raising the second to fourth power and dividing we eliminate $x$
$$ y= \left(\frac{p^3}{q^4}\right)^{\frac{1}{25}} \tag2$$
Plug into either of 1) to find $x$.