Answer to Question #223274 in Algebra for Hanna

Question

If a and b are both positive and unequal, and log_ab + log_ba² =3.

A. b= √a B. b= (√a²)² C. b=a D. b=1/2 a

Answer 1

If $a$ and $b$ are both positive and unequal, and $\log_ab + \log_ba^2 =3.$ It follows that $\log_ab + \frac{\log_aa^2}{\log_ab} =3,$ and hence $\frac{\log^2_ab +2\log_aa}{\log_ab} =3.$ Therefore, $\log^2_ab +2 =3\log_ab,$ which is equivalent to $\log^2_ab-3\log_ab +2 =0,$ and thus to $( \log_ab-1)(\log_ab -2) =0.$ We conclude that $\log_ab=1$ or $\log_ab=2.$ It follows that $\log_ab=\log_aa$ or $\log_ab=2\log_aa=\log_aa^2.$ Consequently, $b=a$ or $b=a^2,\ a>0.$

Taking into account that $a$ and $b$ must be both positive and unequal, we conclude that $b=a^2=((\sqrt{a})^2)^2$ for $a>0.$

Answer: B