7^x-2=4^2-x

To solve the equation \(7^x-2=4^{2-x}\), we need to rewrite both sides of the equation with the same base. We can express \(4^{2-x}\) in terms of \(7\) using the fact that \(4 = 2^2\): \[4^{2-x} = (2^2)^{2-x} = 2^{4-2x} = 2^{2(2-x)} = 2^{4-2x} = 7^{4-2x/(log 7)/(log 2)}\] Now the equation becomes: \[7^x-2 = 7^{4-2x/(log 7)/(log 2)}\] Since the bases are the same, we can equate the exponents: \[x = 4 - 2x/(log 7)/(log 2)\] Now, we solve for \(x\): \[3x = 4\] \[x = 4/3\] Therefore, the solution to the equation is \(x = \frac{4}{3}\).