# Simultaneous equations

#### ssj5harsh

I came across this question once somewhere and I cant seem to get anything:
$$x^y = 9 , y^x = 8$$
Solve the simultaneous equations.
Expressing $x = 9^{1/y}$ and then substituting doesn't help much. I get an equation in one variable, but no direct answer.

Again, taking log on both sides doesn't give anything conclusive.
One answer (at least) is obviously (3,2). How can I solve it mathematically?
I hope someone can help me.

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#### Lucretius

$$x^y = 9 , y^x = 8$$

Try taking

$$\sqrt{x}$$ so you get rid of the y power, making the equation $$y^\sqrt{9}$$$$= 8$$. You can work it out from there.

Of course, you could do $$x^\sqrt{8}$$$$= 9$$, but I wouldn't suggest it, seeing as you will get an approximate answer, versus an integer.

Last edited:

#### VinnyCee

There are 2 equations, and two unknown variables. This means that the equation is solveable by some type of method. We just have to find one. Some knowledge of logaritim and exponential rules comes in handy for this problem.

Taking the natural log of each side allows us to "bring down" any exponentials that may be there:

$$x^{(y)}\,=\,9^{(1)}$$

$$y\,ln\,x\,=\,(1)\,ln\,9$$

Now, divide both sides by natural log of x:

$$y\,=\,\frac{ln\,9}{ln\,x}$$

The algebra is alittle more tricky on the second equation, but this is what I get after doing the same process as above:

$$y\,=\,2^{\frac{3}{x}}$$

Also,

$$\frac{ln\,9}{ln\,x}\,=\,2^{\frac{3}{x}}$$

Now, you can solve it using more algebra (which I did not do!) or just plug these equations into a TI-calculator and see where the graphs intersect or use the solver on the calculator(make sure you are using approximation and not exact) and find the x value where these two equations are equal.

Anyways, x = 3 and then plug that into one of the two equations that were graphed and you get y = 2. Hope this helps.

#### VinnyCee

The second one solved....

$$y^{(x)}\,=\,8^{(1)}$$

First, take ln (natural logarithim) of both sides.

$$ln\left(\,y^{(x)}\,\right)\,=\,ln\left(\,8^{(1)}\,\right)$$

Now you can "bring down" any exponents:

$$(x)\,ln(\,y\,)\,=\,(1)\,ln(\,8\,)$$

Get y on one side only:

$$ln(\,y\,)\,=\,\frac{ln(\,8\,)}{x}$$

Here's an algebra rule that comes in handy, making both sides the exponent of the number "e" (2.271828...) allows you to cancel any terms with $e^{ln(anything)}$:

$$e^{ln(\,y\,)}\,=\,e^{\frac{ln(\,8\,)}{x}}$$

Hence, $e^{ln(anything)}\,=\,anything$ (by "anything", i mean a variable or a constant):

$$y\,=\,e^{\frac{ln(\,8\,)}{x}$$

Here is where I cheated and used the calculator to solve for y (But here's the algebra anyways):

$$e^{ln(\,8\,)\,=\,e^{ln[\,2^{(3)}\,]}$$

$$y\,=\,2^{\frac{3}{x}}$$

Now, solving with the method's above:

$$x\,=\,3,\,\,\,\,\,y\,=\,2$$

Double checking by plugging into original equations:

$$(3)^{(2)}\,=\,9$$

$$(2)^{(3)}\,=\,8$$

#### ssj5harsh

Thank you, VinnyCee. You didn't need to be so descriptive, but I was hoping to find the algabraical way of solving this problem. I don't have a TI-calculator, but I had earlier found the answer with this one:
Graph It!

I also tried the Powertoy calculator which came with Windows XP, though I couldn't figure out how to use it to find the value.

#### HallsofIvy

Homework Helper
Lucretius said:
$$x^y = 9 , y^x = 8$$

Try taking

$$\sqrt{x}$$ so you get rid of the y power, making the equation $$y^\sqrt{9}$$$$= 8$$. You can work it out from there.

Of course, you could do $$x^\sqrt{8}$$$$= 9$$, but I wouldn't suggest it, seeing as you will get an approximate answer, versus an integer.
That would be correct if the first equation were $$x^2= 9$$ rather that $$x^y= 9$$.

#### Lucretius

HallsofIvy said:
That would be correct if the first equation were $$x^2= 9$$ rather that $$x^y= 9$$.
Every time I try to give advice it's wrong . I think I'll just post questions here from now on.

#### dextercioby

Homework Helper
Such transcendental equations cannot be solved analytically,but only numerically or through graph intersection.

Plot implicitely each equation & intersect the graphs.It has 2 solutions.

Daniel.

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