You need to perform the multiplication to the right side, such that:

`2*1156^x - 1 = 34^x`

You need to move all terms to one side, such that:

`2*1156^x - 1 - 34^x = 0`

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

You need to come up with the following replacement, such that:

`34^x = y => 2y^2 - y = 1`

Completing the square to the left side yields:

`2y^2 - y + (1/(2sqrt2))^2 = 1 + 1/(2sqrt2)^2`

`(sqrt2*y - 1/(2sqrt 2))^2 = 9/8`

`sqrt2*y - 1/(2sqrt 2) = +-sqrt(9/8)`

`sqrt2*y = 1/(2sqrt2) +- 3/(2sqrt2)`

`y_(1,2) = (1+-3)/4 => {(y_1 = 1),(y_2 = -1/2):}`

Replacing back `34^x` for y yields:

`34^x = 1 => ln (34^x) = ln 1 => x*ln 34 = 0 => x = 0`

`34^x = -1/2` invalid

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

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.