(1) `log_4 sqrt(x^2+1)=1` Exponentiate both sides with base 4
`4^(log_4 sqrt(x^2+1))=4^1` Exponentiation and logarithms to the same base are inverse operations:
Thus the solutions are `x=sqrt(15)` or `x=-sqrt(15)`
(2) `log(x^2-1)=2+log(x+1)` The understood base is 10, so we exponentiate both sides to base 10:
`10^(log(x^2-1))=10^(2+log(x+1))` Again, exponentiation and logarithms with the same base are inverse functions; also we note that `a^(m+n)=a^m*a^n`
`x=101` or `x=-1`
However, squaring introduced an extraneous solution as `log(x+1)` is undefined when x=-1.
Thus the required solution is `x=101`