1)log base 4 square root of x^2+1 = 1 2) log(x^2-1)= 2+log(x+1) Help!!

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(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`

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