To solve this equation, we first need to use the properties of logarithms to simplify the equation.
First, let's simplify the left side of the equation using the properties of logarithms: Log2 (7x) + Log2(1) = 1 + Log2(3x-5) Log2(7x) + 0 = 1 + Log2(3x-5) Log2(7x) = 1 + Log2(3x-5)
Now, we can use the property of logarithms which states that log_m(a) + log_m(b) = log_m(ab) to combine the terms on the right side of the equation: Log2(7x) = Log2(2) + Log2(3x-5) Log2(7x) = Log2(2(3x-5)) Log2(7x) = Log2(6x - 10)
Since the bases of the logarithms are the same, we can equate the arguments: 7x = 6x - 10 x = -10
Therefore, the solution to the equation Log2 (7x 1) = 1 + log2(3x-5) is x = -10.
To solve this equation, we first need to use the properties of logarithms to simplify the equation.
First, let's simplify the left side of the equation using the properties of logarithms:
Log2 (7x) + Log2(1) = 1 + Log2(3x-5)
Log2(7x) + 0 = 1 + Log2(3x-5)
Log2(7x) = 1 + Log2(3x-5)
Now, we can use the property of logarithms which states that log_m(a) + log_m(b) = log_m(ab) to combine the terms on the right side of the equation:
Log2(7x) = Log2(2) + Log2(3x-5)
Log2(7x) = Log2(2(3x-5))
Log2(7x) = Log2(6x - 10)
Since the bases of the logarithms are the same, we can equate the arguments:
7x = 6x - 10
x = -10
Therefore, the solution to the equation Log2 (7x 1) = 1 + log2(3x-5) is x = -10.