Log6(x+7)=log6(6x-13) Log2(8+3x)=log2(3-x)+1

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To solve the first equation, we can rewrite it as:

x + 7 = 6x - 13

Solving for x, we get:

7 + 13 = 6x -
20 = 5
x = 4

Therefore, the solution for the first equation is x = 4.

For the second equation, we can rewrite it as:

log2(8 + 3x) = log2(3 - x) + 1

Using the rule of logarithms where log_a(x) = log_a(y) is equivalent to x = y, we can rewrite the equation as:

8 + 3x = 2^(3 - x + 1)

8 + 3x = 2^3 * 2^(-x
8 + 3x = 8 / 2^x

Solving for x, we get:

8 + 3x = 8 / 2^
3x = 8 / 2^x -
3x = 8(1 - 2^(-x)
x = 8 / 3(1 - 2^(-x))

Therefore, the solution for the second equation is x = 8 / 3*(1 - 2^(-x)).