How do you solve $log_b(x+2)-log_b4=log_b3x$ ?

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Answer 1 · 2016-09-26T10:32:55

$x=2/11$

$log_b (x+2)-log_b 4=log_b 3x$

$hArrlog_b ((x+2)/4)=log_b 3x$

or $(x+2)/4=3x$

or $x+2=12x$

or $11x=2$ i.e. $x=2/11$

How do you solve log_b(x+2)-log_b4=log_b3xlog_b(x+2)-log_b4=log_b3x?