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How do you solve #log(x-3)+log x=1#?

1 Answer

Bdub · EZ as pi

May 4, 2016

#x=5 #

Explanation:

Use Properties: #log_b(xy)=log_b x+log_by#

#log_bx=y iff b^y=x#

#log (x(x-3))=1# #color(white)(xxxxxx)# [ 1 = log10]

#log(x^2-3x)=log10#

#x^2-3x^1 = 10^1#

#x^2-3x-10=0#

#(x-5)(x+2)=0#

#x=5 or x=-2#

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