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

Answer 1

Isaiah Anthony

#x=10#

#"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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Answer 2

Kennedy Adams

Combine the logarithmic terms, apply logarithmic properties, and solve for x:

[ \log(x) + \log(x-9) = 1 ]

[ \log(x \cdot (x-9)) = 1 ]

[ x \cdot (x-9) = 10 ]

[ x^2 - 9x - 10 = 0 ]

[ (x-10)(x+1) = 0 ]

[ x = 10 \text{ or } x = -1 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.