What are the restrictions on the variable in the equation log(3x-5)-log(x-2)=log(x^2-5)?

1 Answer
Jan 3, 2018

#x=3#

Explanation:

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In the domain of real numbers, you can not have logarithms of zero or negative numbers. As such,

#3x-5 > 0#, #x > 5/3 or 1.666666#

#x-2 > 0#, #x > 2#

#x^2-5 > 0#, #x > |sqrt5| or 2.23607#

We will have to go with #x > |sqrt5|#

Let's solve the equation for #x#:

#log(3x-5)-log(x-2)=log(x^2-5)#

#log((3x-5)/(x-2))=log(x^2-5)#

#(3x-5)/(x-2)=x^2-5#

#x^3-2x^2-8x+15=0#

Using the Rational Zeros Theorem and trying #p/q# values, we find that #3# is a root, As such,

#(x-3)(x^2+x-5)=0#

We get the following three answers for #x#:

#x=3, (-1+sqrt21)/2, (-1-sqrt21)/2#

Based on #x > |sqrt5|# the only acceptable answer is #x=3# as the solution to the equation because the middle one is about #1.79# and the last one is negative.