To solve this logarithmic equation, we will first use the properties of logarithms to rewrite the equation in a simpler form.
log^36(2x-7)^2 = log^6(14-x)
(2x-7)^2 = 14-x
4x^2 - 28x + 49 = 14 - x
4x^2 - 27x + 35 = 0
(4x - 5)(x - 7) = 0
4x - 5 = 0 -> 4x = 5 -> x = 5/4x - 7 = 0 -> x = 7
Therefore, the solutions to the equation are x = 5/4 and x = 7.
To solve this logarithmic equation, we will first use the properties of logarithms to rewrite the equation in a simpler form.
Apply the power rule of logarithms, which states that loga(b^c) = c*loga(b):log^36(2x-7)^2 = log^6(14-x)
Apply the rule of logarithms that states if loga(b) = loga(c), then b = c:(2x-7)^2 = 14-x
Square the left side of the equation:4x^2 - 28x + 49 = 14 - x
Rearrange the terms:4x^2 - 27x + 35 = 0
Factor the quadratic equation:(4x - 5)(x - 7) = 0
Set each factor to zero and solve for x:4x - 5 = 0 -> 4x = 5 -> x = 5/4
x - 7 = 0 -> x = 7
Therefore, the solutions to the equation are x = 5/4 and x = 7.