Question

log5(x+2)+log5(2x-1)=2??

Answer 1

do 5 is the base of the given log?

Answer 2

yes

Answer 3

\[\log_{5}(x+2)+\log_{5}(2x-1)=2\]
\[\log_{5}[(x+2)(2x-1)]=2\]
i.e. \[(x+2)(2x-1)=5^2\]
i.e. (x+2)(2x-1)=25
i.e\[ (x+2)(2x-1)=25 \rightarrow 2x^2-x+4x-2=25 \rightarrow 2x^2+3x-27=0\]
i.e. 2x^2+9x-6x-27=0
i.e. x(2x+9)-3(2x+9)=0
i.e. (x-3)(2x+9)=0
i.e. (x-3)=0 or (2x+9)=0
i.e. x=3 or x=-9/2
Since x=-9/2 is negative
therefore x=3 is the ans