Practice Exponential Equations 3

MarkFL

La Villa Strangiato
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,983
4,504
113
St. Augustine
Observe that:

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

Take it away!!! :)
 
  • Like
Reactions: pre-trip_rapture

MarkFL

La Villa Strangiato
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,983
4,504
113
St. Augustine
Because a power of 1 makes the equation true, when we observe that:

\(\displaystyle 5+a+5-a=10\)
 

MarkFL

La Villa Strangiato
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,983
4,504
113
St. Augustine
And so we write:

\(\displaystyle x^2-5x+5=1\)

\(\displaystyle x^2-5x+4=0\)

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

\(\displaystyle x\in\{1,4\}\)
 

MarkFL

La Villa Strangiato
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,983
4,504
113
St. Augustine
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:

\(\displaystyle x\in\{1,2,3,4\}\)
 
  • Like
Reactions: pre-trip_rapture
Jan 15, 2020
659
7
18
54
Bronx, NY
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
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,983
4,504
113
St. Augustine
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.
 
  • Like
Reactions: pre-trip_rapture
Jan 15, 2020
659
7
18
54
Bronx, NY
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
Staff member
Administrator
Moderator
Math Helper
Jan 25, 2018
3,983
4,504
113
St. Augustine
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.
 
  • Like
Reactions: pre-trip_rapture
Jan 15, 2020
659
7
18
54
Bronx, NY
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.