Take the exponential on both sides:
e ln ( x ) + ln ( x + 3 ) = e ln ( x ) ⋅ e ln ( x + 3 ) = x ( x + 3 ) {\displaystyle e^{\ln(x)+\ln(x+3)}=e^{\ln(x)}\cdot e^{\ln(x+3)}=x(x+3)} and e ln ( 20 − 5 x ) = 20 − 5 x {\displaystyle e^{\ln(20-5x)}=20-5x} .
We then solve the quadratic equation:
x ( x + 3 ) = 20 − 5 x {\displaystyle x(x+3)=20-5x}
( x 2 + 3 x ) − ( 20 − 5 x ) = x 2 + 8 x − 20 = 0 {\displaystyle (x^{2}+3x)-(20-5x)=x^{2}+8x-20=0}
It has solutions x = − 10 {\displaystyle x=-10} and x = 2 {\displaystyle x=2} .
We note that x = − 10 {\displaystyle x=-10} is not in the domain of ln ( x ) {\displaystyle \ln(x)} .
Therefore the equation only has one solution: x = 2 {\displaystyle x=2} .
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