Show that $\log_{25}(81x^2)=\log_5 9+\log_5 x$ for $x0$. Hence, solve the equation \(\log_5\sqr

Question

Show that $\log_{25}(81x^2)=\log_5 9+\log_5 x$ for $x>0$. Hence, solve the equation $\log_5\sqrt[3]{x}=\log_{25}(81x^2)-\log_5 4$, where $x>0$.

Accepted Answer

(b) $x=\frac8{27}$

IB AA HL May 2026 P1 TZA Q4