# What are the solutions of the exponential 2*2025^x-(3*15)^x=1?

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To find the solution of 2*2-25^x-(3*15)^x = 1.

We know that 2025 = (3*15)^2 = 45^2.

So 2025^x= (3* 15)^2x = 45^2x.

Therefore the given equation becomes:

2* 45^2x - (45)^x -1 = 0 1.

2t^2-t-1 = 0, where t = 45^x.

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

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

t -1 = 0 gives 45^x = 1, or 45^x= = 1= 45^0. So x = 0.

2t+1 =0 gives t = -1/2, or 45^x = -1/2 which has no real solution.

**Therefore x= 0 is the solution.**

We notice that 2025=45^2! (1)

We also notice that the term (3*15)^x = 45^x

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

2025^x = 45^2x

We'll re-write the equation, moving all terms to the left side:

2*45^2x - 45^x - 1=0

We'll substitute 45^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 determine the x values:

45^x=1

45^x = 45^0

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

x = 0

45^x = -1/2

The exponential 45^x is always positive, for any value of x, so, 45^x = -1/2 is undefined.

**The equation has only one solution, x = 0.**