# Solve for x the equation 289^x - 17^(x+1) + 16 = 0?

*print*Print*list*Cite

### 3 Answers

The equation to be solved is 289^x - 17^(x+1) + 16 = 0

289^x - 17^(x+1) + 16 = 0

=> 17^2^x - 17*17^x + 16 = 0

Let 17^x = y

=> y^2 - 17y + 16 = 0

=> y^2 - 16y - y + 16 = 0

=> y(y - 16) - 1(y - 16) = 0

=> (y - 1)(y - 16) = 0

=> y = 1 and y = 16

17^x = 1 and 17^x = 16

=> x = 0 and x = log 16/log 17

**The solution of the equation is x = 0 and x = log 16/log 17**

We notice that 289 is a power of 17.

289 = 17^2

We'll raise both sides by x;

289^x = 17^2x

We'll re-write the equation:

17^2x - 17*17^x + 16 = 0

We'll substitute 17^x = t:

t^2 - 17t + 16 = 0

We'll apply the quadratic:

t1 = [17+sqrt(289-64)]/2

t1 = (17+15)/2

t1=16

t2=(17-15)/2

t2=1

17^x = t1 <=> 17^x =16

Since the bases are not matching, we'll take natural logarithms both sides:

x*ln17 = ln16

x = ln16/ln17

x = 2.7725/2.8332

x=0.9785

17^x = t2

17^x = 1 <=> x = 0

**The solutions of the equation are {0 ; 0.9785}.**

We have to solve the equation 289^x - 17^(x+1) + 16 = 0. Notice that 289 = 17^2.

Rewrite the equation that is given in the following way to arrive at a quadratic equation.

289^x - 17^(x+1) + 16 = 0

(17^2)^x - 17^(x+1) + 16 = 0

(17^x)^2 - 17^x*17 + 16 = 0

Let 17^x = y

y^2 - 17y + 16 = 0

y^2 - 16y - y + 16 = 0

y(y - 16) - 1(y - 16) = 0

(y - 1)(y - 16) = 0

y = 1 and y = 16

As y = 17^x

17^x = 1 and 17^x = 16

x = 0 and x = `(ln 16)/(ln 17)`

The equation has two solutions x = 0 and x = `(ln 16)/(ln 17)`