1) 8^2y = 52^4y+3

2) 2^3(2y) = 2^26(4y+3)
3) 6y = 104y+78
4) -98y = 78
5) y= -0.796

1) In order to solve the equation 8^(2y) = 52^(4y+3), we can start by expressing both sides of the equation with the same base. Since both 8 and 52 are powers of 2, we can rewrite it as (2^3)^(2y) = (2^2)^((4y+3)log2(52)).

Next, we can simplify the equation by using the power of a power rule, which states that (a^m)^n = a^(m*n). Applying this rule, we get 2^(6y) = 2^((8y+6)log2(52)).

To find a solution, we equate the exponents: 6y = (8y+6)log2(52).

Next, we isolate the variable y. We can divide both sides of the equation by log2(52) to get rid of the logarithm term: (6y)/(log2(52)) = 8y + 6.

Further simplifying, we can subtract 8y from both sides: -2y = 6.

Finally, we divide both sides by -2 to solve for y: y = -3.

2) The equation 2^(3(2y)) = 2^(26(4y+3)) is set up similar to the previous equation. We start by using the power of a product rule, which states that (a^m)^n = a^(m*n). Applying this rule, we can simplify the left side of the equation to 2^(6y).

For the right side, we can use the power of a sum rule, which states that a^(m+n) = a^m * a^n. Applying this rule, we simplify the right side of the equation to 2^(104y+78).

Now, we have the equation 2^(6y) = 2^(104y+78).

To find a solution, we can equate the exponents: 6y = 104y + 78.

Next, we isolate the variable y. By subtracting 104y from both sides, we get -98y = 78.

Finally, we divide both sides by -98 to solve for y, yielding y = -0.796.

3) The equation 6y = 104y + 78 can be solved by isolating the variable y.

First, we can subtract 104y from both sides of the equation: 6y - 104y = 78.

Simplifying further, we get -98y = 78.

Finally, dividing both sides by -98 gives us y = -0.796.

4) In the equation -98y = 78, we can isolate the variable y by dividing both sides of the equation by -98: y = 78 / -98.

Simplifying further, y = -0.796.

5) There is no equation given, so the value y = -0.796 is already the solution.