1968 AHSME Problems/Problem 23

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Problem

If all the logarithms are real numbers, the equality $log(x+3)+log(x-1)=log(x^2-2x-3)$ is satisfied for:

$\text{(A) all real values of }x \quad\\ \text{(B) no real values of } x\quad\\ \text{(C) all real values of } x \text{ except } x=0\quad\\ \text{(D) no real values of } x \text{ except } x=0\quad\\ \text{(E) all real values of } x \text{ except } x=1$


Solution

$\fbox{B}$

Solution 2

From the given we have \[\log(x+3)+\log(x-1)=\log(x^2-2x-3)\] \[\log(x^2+2x-3)=\log(x^2-2x-3)\] \[x^2+2x-3=x^2-2x-3\] \[x=0\] However substituing into $\log(x-1)$ gets a negative argument, which is impossible $\boxed{D}$.

~ Nafer

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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