# PracticeExponential Equations 3

#### MarkFL

##### La Villa Strangiato
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Observe that:

$$\displaystyle 10=(5+2\sqrt{6})^1+(5-2\sqrt{6})^1$$

Take it away!!!

pre-trip_rapture

#### pre-trip_rapture

##### Member
Observe that:

$$\displaystyle 10=(5+2\sqrt{6})^1+(5-2\sqrt{6})^1$$

Take it away!!!
How did you get a power of 1?

#### MarkFL

##### La Villa Strangiato
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Because a power of 1 makes the equation true, when we observe that:

$$\displaystyle 5+a+5-a=10$$

#### MarkFL

##### La Villa Strangiato
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And so we write:

$$\displaystyle x^2-5x+5=1$$

$$\displaystyle x^2-5x+4=0$$

$$\displaystyle (x-4)(x-1)=0$$

$$\displaystyle x\in\{1,4\}$$

#### pre-trip_rapture

##### Member
Observe that:

$$\displaystyle 10=(5+2\sqrt{6})^1+(5-2\sqrt{6})^1$$

Take it away!!!

#### MarkFL

##### La Villa Strangiato
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It appears we missed a real solution. Let's look at:

$$\displaystyle \frac{1}{5\pm2\sqrt{6}}\cdot\frac{5\mp2\sqrt{6}}{5\mp2\sqrt{6}}=5\mp2\sqrt{6}$$

So, we see the two bases are each others' reciprocals. Let's begin by writing:

$$\displaystyle (5+2\sqrt{6})^u+(5-2\sqrt{6})^u=10$$

Using our discovery above, we can now write:

$$\displaystyle (5+2\sqrt{6})^u+(5+2\sqrt{6})^{-u}=10$$

If we multiply through by $$v=(5+2\sqrt{6})^u$$ then we get (after arranging in standard form):

$$\displaystyle v^2-10v+1=0$$

$$\displaystyle v=5\pm2\sqrt{6}$$

This means:

$$\displaystyle (5+2\sqrt{6})^u=5\pm2\sqrt{6}$$

And using our initial result above, this means:

$$\displaystyle u=x^2-5x+5=\pm1$$

We've previously solved the positive case, so now the negative:

$$\displaystyle x^2-5x+5=-1$$

$$\displaystyle x^2-5x+6=0$$

$$\displaystyle (x-2)(x-3)=0$$

And so the complete set of real solutions is:

$$\displaystyle x\in\{1,2,3,4\}$$

pre-trip_rapture

#### pre-trip_rapture

##### Member
It appears we missed a real solution. Let's look at:

$$\displaystyle \frac{1}{5\pm2\sqrt{6}}\cdot\frac{5\mp2\sqrt{6}}{5\mp2\sqrt{6}}=5\mp2\sqrt{6}$$

So, we see the two bases are each others' reciprocals. Let's begin by writing:

$$\displaystyle (5+2\sqrt{6})^u+(5-2\sqrt{6})^u=10$$

Using our discovery above, we can now write:

$$\displaystyle (5+2\sqrt{6})^u+(5+2\sqrt{6})^{-u}=10$$

If we multiply through by $$v=(5+2\sqrt{6})^u$$ then we get (after arranging in standard form):

$$\displaystyle v^2-10v+1=0$$

$$\displaystyle v=5\pm2\sqrt{6}$$

This means:

$$\displaystyle (5+2\sqrt{6})^u=5\pm2\sqrt{6}$$

And using our initial result above, this means:

$$\displaystyle u=x^2-5x+5=\pm1$$

We've previously solved the positive case, so now the negative:

$$\displaystyle x^2-5x+5=-1$$

$$\displaystyle x^2-5x+6=0$$

$$\displaystyle (x-2)(x-3)=0$$

And so the complete set of real solutions is:

$$\displaystyle x\in\{1,2,3,4\}$$
Nicely done as always. Are you saying that my simply picture answer is not even close to proving that the LHS = RHS?

#### MarkFL

##### La Villa Strangiato
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When the exponent is 1, the equation is true, but we missed the fact that when the exponent is -1, it is also true. Both values of the exponent lead to 2 values of $$x$$ which make the equation true.

pre-trip_rapture

#### pre-trip_rapture

##### Member
When the exponent is 1, the equation is true, but we missed the fact that when the exponent is -1, it is also true. Both values of the exponent lead to 2 values of $$x$$ which make the equation true.
I do not feel so bad when you said "we missed the fact" in your reply.

#### MarkFL

##### La Villa Strangiato
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Yes, I "eyeballed" it and saw the obvious value for the exponent, but in doing so missed the not so obvious value. I found it when I revisited this problem looking for a more algebraic way to solve the equation.

pre-trip_rapture

#### pre-trip_rapture

##### Member
Yes, I "eyeballed" it and saw the obvious value for the exponent, but in doing so missed the not so obvious value. I found it when I revisited this problem looking for a more algebraic way to solve the equation.
Now it is solved for all to see.