Log question help please

solve for x

a) 7^(2x+1) = 5

b) 4^(x) = 2^(x-1)

a^b = e^(b ln a)

a) 7^(2x+1) = 7^(2x) * 7^1 = (7²)^x * 7 = 5
49^x = 5/7
e^(x ln(49)) = 5/7
x ln 49 = ln(5/7) = ln 5 - ln 7
x = (ln 5 - ln 7) / ln 49

b) 2^(x-1) = 2^x * 2^(-1) = 2^x / 2 = 4^x
e^(x ln 2) / 2 = e^(x ln 4)
ln(e^(x ln 2) / 2) = ln(e^(x ln 4))
ln(e^(x ln 2)) - ln 2 = ln(e^(x ln 4))
x ln 2 - ln 2 = x ln 4
x (ln 2 - ln 4) = ln 2
x = ln 2 / (ln 2 - ln 4)
x = ln 2 / (ln 2/4)
x = ln 2 / ln (1/2)
x = ln 2 / (ln 1 - ln 2)
x = ln 2 / (0 - ln 2)
x = ln 2 / -ln 2
x = -1

a) (2x+1)=log 5/log 7=0.827
b) 2x=x-1
x=-1
God bless you.