How do you solve $log (x + 1) = 2 + log x$ $log(x+1) = 2 + log x$ ?

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Answer 1 · 2015-12-01T22:03:31

$x = \frac{1}{99}$ $x=1/99$

Know that: ${(loga-logb=log (a/b)),(loga=b<=>a=10^b):}$

$log(x+1)=2+logx$

$log(x+1)-logx=2$

$log((x+1)/x)=2$

$(x+1)/x=10^2$

$x+1=100x$

$x=1/99$

How do you solve log(x+1) = 2 + log xlog(x+1)=2+logxlog(x+1) = 2 + log x?