25^(x+1) +49 *5^x -2=0

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To solve this equation for x, we need to use algebraic manipulation.

Let's rewrite the equation as:

25^(x+1) + 49*5^x - 2 = 0

Now, we know that 25 = 5^2, so we can rewrite the equation as:

(5^2)^(x+1) + 49*5^x - 2 = 0

Following the rules of exponents, we can simplify:

5^(2x+2) + 49*5^x - 2 = 0

Now, we have a common base of 5, so we can combine the terms:

5^(2x+2) + 50*5^x - 2 = 0

Now, let's rewrite 50 as 5*10:

5^(2x+2) + 5105^x - 2 = 0

Combining like terms:

5^(2x+2) + 50*5^x - 2 = 0

Let's rewrite 50 as 5^2:

5^(2x+2) + 5^2*5^x - 2 = 0

Now, using the rules of exponents, we can add the powers:

5^(2x+2+x) - 2 = 0

5^(3x+2) - 2 = 0

Now, we can add 2 to both sides to isolate the exponential term:

5^(3x+2) = 2

Taking the logarithm of both sides with base 5:

log5(5^(3x+2)) = log5(2)

Using the property of logarithms:

3x + 2 = log5(2)

Now, we can solve for x:

3x = log5(2) - 2

x = (log5(2) - 2) / 3

This is the solution to the equation 25^(x+1) + 49*5^x - 2 = 0.