# Find the roots of the equation 2*1156^x-1=(2*17)^x.

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### 2 Answers

We notice that 1156=34^2! (1)

We also notice that the right side term (2*17)^x = 34^x

We'll raise to x both sides the equality (1):

1156^x = 34^2x

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

2*(34^2)^x - 34^x - 1=0

We'll substitute 34^x by another variable, t.

2*t^2 - t - 1=0

We'll apply the quadratic formula:

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**

**Now, we'll find the x values:**

34^x=1

34^x = 34^0

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

x = 0

34^x = -1/2

The exponential 34^x is always positive, for any value of x, so, we'll reject the second solution.

**The equation has just one solution, namely x = 0.**

To solve 2*1156^x-1 = (2*17)^x.

We see that 1156 = 17*68 = 17^2*2^2 = {(17*2)^x}^2

Therefore we put (17*2)^x = t.

Then 2*1156^x = 2t^2.

Therefore the given equation becomes 2t^2-1 = t, where t = (17*2)^x. Subtracting t from both sides, we get:

2t^2 -t -1 = 0.

2t-2t +t-1 = 0.

2t(t-1) +1(t-1) = 0.

(t-1)(2t+1) = 0.

t-1 = 0 gives t = 1. So t = (17*2)^x = 1 = (17*2)^0.

So x=0. 2t+1 = 0 does not give real solution.

Therefore x = 0.