Can somebody help me with this problem? ?

Question

3^x + 3^-x = 6

Solve the above equation algebraically.

Lower value: x =

Higher value: x =

Please show solutions.

Answer 1

3^x + 3^(-x) = 6

Let Y = 3^x

So now ...

Y + 1/Y = 6

Multiply both side by Y

Y(Y + 1/Y) = 6Y

Y^2 + Y/Y = 6Y

Y^2 + 1 = 6Y

Put into simple quadratic equation form (force 1 side = 0)

Y^2 - 6Y +1 = 0

It doesn't factor, let's complete the square so it will factor

Y^2 - 6Y + 1 +8 = 8

Y^2 -6y + 9 = 8

(Y-3)^2 = 8

Take the square root of both sides

Y - 3 = √8

Y = 3 ± √8

Y = 3 + √8 and Y = 3 - √8

3^X = 3 + √8 3^x = 3 - √8

3^x = 3 + 2.828 3^x = 3 - 2.828

3^x = 5.828 3^x = 1.72

Extrapolate the values for x from tables:

x = ± 1.608 (approximately)

Check the answer:

3^x + 3^-x = ...

3^1.608 + 3^-1.608 = ...

5.850 + .017 = 6.02

Since the original equation raises the same value (3) to both positive and negative values the second answer -1.608 will yield the same answer and we have no need to check that one.

Correct, .02 difference for the rounding in the

approximation of x

Answer 2

3^x + 3^-x = 6

-> 3^x + 1/3^x = 6

Let 3^x = y

So, y + 1/y = 6

Multiply by y

-> y² + 1 = 6y

-> y² - 6y + 1 = 0

Solving the quadratic equation for y,

y = 5.828 & y = 0.172 (approximately)

Now we have y = 3^x

So 3^x = 5.828 & 3^x = 0.172

For 3^x = 5.828

Taking log on both sides

x log 3 = log 5.828

-> x = log 5.828 / log 3 = 1.604 (approximately)

Similarly, for 3^x = 0.172

x log 3 = log 0.172

-> x = log 0.172 / log 3 = -1.602 (approximately)

Therefore,

Lower value: x = -1.602 (approximately)

Higher value: x = 1.604 (approximately)

Answer 3

I think this is the answer:

3^x+3^-x=6

3^x+1/3^x=6

We'll place:

3^x=X

in the equation:

X+1/X=6  /*X

X^2+1=6

X^2=5

X=√5  and X=-√5