# Find the solutions of the equation 2*16^x = 4^x + 1 .Find the solutions of the equation 2*16^x = 4^x + 1 .

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You need to convert the given exponential equation into a quartic equation, such that:

`2^x = y => {(4^x = y^2),(16^x = y^4):}`

Changing the variable yields:

`2y^4 = y^2 + 1 => 2y^4 - y^2 - 1 = 0`

`y^4 + y^4 - y^2 - 1 = 0 => (y^4 - 1) + (y^4 - y^2) = 0`

`(y^2 - 1)(y^2 + 1) + y^2(y^2 - 1) = 0`

Factoring out `(y^2 - 1)` yields:

`(y^2 - 1)(y^2 + 1 + y^2) = 0 => {(y^2 - 1 = 0),(2y^2 + 1 = 0):}`

`{(y^2 = 1),(y^2 = -1/2 !in R):}=> y_(1,2) = +-1`

Replacing back `2^x` for y yields:

`2^x = 1 => ln 2^x = ln 1 => x ln 2 = 0 => x = 0`

`2^x = -1` invalid

**Hence, evaluting the solution to the given equation yields **`x = 0.`

We notice that 16=4^2!

We'll re-write the equation in this manner:

2*(4^2)^x - 4^x - 1=0

We'll substitute 4^x by another variable, t.

2*t^2 - t - 1=0

t1=[1+sqrt (1+4*2)]/4

t1=[1+sqrt (9)]/4

t1=(1+3)/4

t1=1

t2=[1-sqrt (1+4*2)]/4

t2=(1-3)/4

t2=-1/2

We didn't find the values of x, yet!

4^x=1

4^x=4^0

Since the bases are matching, we'll apply one to one property:

x = 0

4^x=-1/2

The exponential 4^x is always positive, for any value of x, so, we'll reject the second solution.

The equation has just one solution. The only solution is x= 0.