## Algebra 2 Common Core

Error in line 3 (dropping logs around values). Correct answer: $x=16$
Notice that in the third line, the "logs" were dropped from the equation. We can do this if the two logs have the same base. However, in our case the bases ($2$ and $3$) do not match. Thus we cannot drop the logs from the equation. Instead we proceed as follows: $\log_2 x=\log_3 9^2$ $\log_2 x=\log_3 (3^2)^2$ $\log_2 x=\log_3 3^4$ Recall the power property of logarithms (pg. 462): $\log_b{m^n}=n\log_b{m}$ Applying this property to our last equation, we get: $\log_2 x=4\log_3 3$ $\log_2 x=4\times 1$ $\log_2 x=4$ We also used the fact that $\log_{3}{3}=1$ (because $3^1=3$). Next, recall the definition of a logarithm (pg. 451): $\log_{b}{x}=y$ iff $b^y=x$ Applying this to our last equation, we get: $\log_2 x=4$ $2^4=x$ $16=x$ $x=16$