Given that $\log_3(x^2 - 2x) = 1$, find the possible values of $x$.
A$x = 3$ or $x = -1$CORRECT
B$x = 3$ or $x = 1$
C$x = -3$ or $x = 1$
D$x = 6$ or $x = -3$
GCEMathematicsIndices, Logarithms and Surds2021
AI
Toaster Teacher
Why the answer is A, and why the others tempt you.
$\log_3(x^2 - 2x) = 1$ implies $x^2 - 2x = 3^1 = 3$, so $x^2 - 2x - 3 = 0$, which factors as $(x-3)(x+1) = 0$, giving $x = 3$ or $x = -1$. Both values must be checked: $x=3$ gives $\log_3 3 = 1$ ✓ and $x=-1$ gives $\log_3(1+2)=\log_3 3=1$ ✓, so both are valid.
Want this in Pidgin, Yoruba, Igbo or Hausa? Sign up free →
Practice more Mathematics questions
GCE Mathematics has thousands more questions like this — with Worked answers on every one.