How do you solve for x?: #log(x) + log(x-9) = 1#

1 Answer
Jim G.
Apr 17, 2018

#x=10#

Explanation:

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx+logy=log(xy)#

#•color(white)(x)log_b x=nhArrx=b^n#

#"note that "logx-=log_(10)x#

#rArrlog_(10)x(x-9)=1#

#rArrx^2-9x=10^1=10#

#rArrx^2-9x-10=0larrcolor(blue)"in standard form"#

#rArr(x-10)(x+1)=0#

#rArrx=10" or "x=-1#

#"note that "x>0" and "x-9>0#

#rArrx=-1" is invalid"#

#rArrx=10" is the solution"#

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