Logarithm Problem? Algebra 2? I have a test on this tomorrow!?

Question

so i was studying for this algebra 2 test i have tomorrow and i came across this problem.
2log(base-a)x + log(base-a)2 = log(base-a)(5x+3)

Answer 1

2log(base-a)x + log(base-a)2 = log(base-a)(5x+3)

Using logarthamic property alogb=log(b^a)

log(base-a)x^2 + log(base-a)2 = log(base-a)(5x+3)

Using addition property of logarthims loga+logb=logab

log(base-a)(2x^2)  = log(base-a)(5x+3)

2x^2=5x+3

2x^2-5x-3=0

Solving above equation we get

2x^2-6x+x-3=0

2x(x-3)+1(x-3)=0

2x+1=0 or x-3=0

x=-1/2 or x=3

Comments

oh man ,thank u very much good approach.