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 \(x = \frac{4}{3}\).