# Find a if a + a^-1 = 2. Find x if 5^x = a

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a+a^-1 = 2 . To find the x in 5^x=a.

a+1/a =2

a^2+1=2a

a^2-2a+1 = 0

(a-1)^2 = 0

so a -1 = 0

a= 1.

Therefore to find x in 5^x = a = 1

5^x = 1 = 5^0

Equating the powers, we get:

x = 0.

We'll re-write the equation, using the property of negative exponents:

a^-1 = 1/a

The equation will become:

a + 1/a = 2

a^2 + 1 = 2a

We'll subtract 2a both sides:

a^2 - 2a + 1 = 0

The expression is the expanding of the square (a-1)^2.

(a-1)^2 = 0

**a1= a2 = 1**

Now, we'll determine x knowing that a = 5^x.

1 = 5^x

We'll write 1 as the power of 5^0:

5^0 = 5^x

Since the bases are matching, we'll use the one to one property and we'll get:

**x = 0**

We are given: a + a^-1 = 2

a + a^-1 = 2

Multiply both sides by a^1

=> a^2 + a^0 = 2a^1

=> a^2 + 1 = 2a

subtract 2a from both sides

=> a^2 - 2a + 1 = 0

(a - 1)^2 = a^2 -2a + 1

=> (a -1)^2 =0

=> a-1 = 0

=> a=1

So the value of a is 1.

If 5^x = a =1

=> x= 0 as any number to the power 0 is 1

**So x is equal to 0.**