# What is x if 9^x −10*3^(x−1) +1= 0?

### 3 Answers | Add Yours

The equation 9^x -10*3^(x-1) +1= 0 has to be solved for x.

9^x - 10*3^(x-1) +1= 0

Use the property x^(a - b) = x^a/x^b

3^(x - 1) = 3^x/3

9^x = (3^2)^x, Using the property (x^a)^b = (x^b)^a gives 9^x = (3^x)^2

The equation can now be written as:

(3^x)^2 - 10*3^x/3 + 1 = 0

Let y = 3^x

3*y^2 - 10y + 3 = 0

3y^2 - 9y - y + 3 = 0

3y(y - 3) - 1(y - 3) = 0

(3y - 1)(y - 3) = 0

3y - 1 = 0

y = 1/3

As y = 3^x, x = -1

y - 3 = 0

y = 3

As y = 3^x, x = 1

The solution of the equation 9^x -10*3^(x-1) +1= 0 is x = 1 and x = -1

Using the laws of exponents

9^x=(3^2)^x=3^2x

3^(x-1)=3^x / 3^1=(3^x)/3

The equation becomes

3^2x - (10/3)*3^x+1 = 0

Let 3^X=a

a^2 - (10/3)a +1=0

Apply the quadatic formula

a=(10/3 +/- sqrt of (100/9 - 4*1*1))/2

=(6 or 2/3)/2

=3 or 1/3

3=3^x

x=1

1/3= 3 ^x

3^-1=3^x

x=-1

The solutions are -1 or 1

We'll create matching bases for all the exponential terms.

9^x = 3^2x

3^(x-1) = 3^x/3

We'll re-write the equation:

3^2x - 10*3^x/3 + 1 = 0

We'll replace 3^x by t;

t^2 - 10t/3 + 1 = 0

3t^2 - 10t + 3 = 0

We'll apply the quadratic formula:

t1 = [10+sqrt(100 - 36)]/6

t1 = (10+8)/6

t1 = 3

t2 = 1/3

t1 = 3^x => 3 = 3^x

Since the bases of exponentials are matching, we'll apply one to one rule:

x = 1

t2 = 3^x => 1/3 = 3^x => 3^-1 = 3^x

x = -1

**The soutions of the equation are {-1 ; 1}.**