Please help me with this logarithm equation:
Thanks - solve in terms of x.
$\displaystyle \log_2\left(2^{2x}-56\right) - x=0~\implies~\log_2\left(2^{2x}-56\right) = x$
Now use the base 2 with both sides of the equation:
$\displaystyle 2^{2x}-56 = 2^x$
Use substitution $\displaystyle 2^x = y$ that means y > 0.
You'll get:
$\displaystyle y^2-y-56 = 0~\implies~y = -7~\vee~y=8$
You only can use the second solution:
$\displaystyle 8=2^x~\implies~x=3$