Logarithmic Equation and Domain Restrictions
An equation is given as log_2(x + 3) + log_2(x - 1) = 3. A student combines the logarithms, solves the resulting quadratic, and obtains x = -1 + 2 sqrt(3) and x = -1 - 2 sqrt(3). They then solve log_2(x + 3) + log_2(x - 1) < 3 and write the solution as -1 - 2 sqrt(3) < x < -1 + 2 sqrt(3). A classmate objects that the original expression is not defined for x <= 1, so the lower endpoint cannot be part of the answer. The student must decide how to present the solution set, explain why one algebraic root is rejected in the equation but still appears as an endpoint in the quadratic inequality, and avoid losing marks for including values outside the domain. No calculator is allowed; exact surds are required.
1. Solve the equation, showing the domain restrictions and rejecting any invalid solution.
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The logarithms require x+3>0 and x-1>0, so the domain is x>1.
Combine: log_2((x+3)(x-1))=3, so (x+3)(x-1)=2^3=8. Hence x^2+2x-3=8, giving x^2+2x-11=0.
The quadratic formula gives x=-1±sqrt(12)=-1±2sqrt(3). Now -1-2sqrt(3) is less than 1, so it is outside the domain and must be rejected. The only solution is x=-1+2sqrt(3). This is approximately 2.46, which is greater than 1, so it is valid.
2. Solve the inequality, giving the final solution set in exact form and explaining the role of the domain.
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The domain is still x>1. Because log_2 is an increasing function, on this domain log_2((x+3)(x-1))<3 is equivalent to (x+3)(x-1)<8.
This gives x^2+2x-11<0, so (x+1)^2<12, i.e. -1-2sqrt(3)<x<-1+2sqrt(3). Intersect with x>1: the lower root is negative and is discarded.
Hence 1<x<-1+2sqrt(3). Both endpoints are excluded: x=1 is outside the domain, and x=-1+2sqrt(3) makes the inequality equal rather than less than. Final answer: x in (1, -1+2sqrt(3)).
3. A student writes: 'The inequality solution is -1 - 2sqrt(3) < x < -1 + 2sqrt(3) because the quadratic is negative between its roots.' Identify the error and explain how a sketch of the logarithmic expression would prevent it.
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The error is treating the quadratic inequality as if it were the original logarithmic inequality. The original expression is defined only for x>1, so any interval obtained from the quadratic must be intersected with x>1.
The lower endpoint -1-2sqrt(3) comes from the quadratic but lies outside the domain; it is not part of the solution set. A sketch of y=log_2(x+3)+log_2(x-1) would show the graph existing only to the right of x=1, so the solution cannot include values below 1. The correct solution is 1<x<-1+2sqrt(3).