(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:

`sqrt(x^2+1)=4`

`x^2+1=16`

`x^2=15`

`x=+-sqrt(15)`

**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^2-1=10^2*(x+1)`

`x^2-1=100x+100`

`x^2-100x-101=0`

`(x-101)(x+1)=0`

`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`

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