Problem:
Find the value of x if 7^(2x+3) = 1.

Explanation:
To find the value of x in the given equation, we need to manipulate the equation using the properties of exponents and solve for x.

Properties of Exponents:
1. a^0 = 1, where a is any non-zero number.
2. a^m * a^n = a^(m+n), where a is any non-zero number and m and n are any real numbers.
3. (a^m)^n = a^(m*n), where a is any non-zero number and m and n are any real numbers.
4. a^(-n) = 1/a^n, where a is any non-zero number and n is any real number.

Manipulating the Equation:
1. We start by applying the property of exponents in reverse to isolate the base 7:
7^(2x+3) = 1
Taking the logarithm of both sides with base 7:
log7(7^(2x+3)) = log7(1)
Using the property of logarithms to bring down the exponent:
(2x+3) * log7(7) = 0
Simplifying the logarithm:
(2x+3) * 1 = 0
2x + 3 = 0

2. Now, we solve the equation for x:
2x + 3 = 0
Subtracting 3 from both sides:
2x = -3
Dividing both sides by 2:
x = -3/2

Final Answer:
The value of x in the equation 7^(2x+3) = 1 is x = -3/2.

-3/2