7*(2^(3x-5))=53 solve for x
\(7\times2^{3x-5} = 53\\ 2^{3x-5}=\dfrac{53}{7}\\ \log_2 (2^{3x-5})=\log_2\dfrac{53}{7}\\ 3x - 5 = \log_2\dfrac{53}{7}\\ x = \dfrac{\log_2\dfrac{53}{7}+5}{3}\)
Which is 7.920565532505595
