Question

Log base b of x = 2 - a + log base b ((a^2 * b^a)/b^2)

Answer 1

\[\log_b x= 2-a + \log_b \frac{a^2b^a}{b^2}\]
Let's bring the log term on the right hand side to left hand side
\[\log_b x- \log_b \frac{a^2b^a}{b^2}= 2-a \]
We know
\[\log_b y- \log_b z= \log _b \frac{y}{z}\]
so we get
\[\log_b \frac{xb^2}{a^2b^a}= 2-a\]
raise both sides to the power of b
\[\large b^{\log_b \frac{xb^2}{a^2b^a}}= b^{2-a}\]
We know
\[\large b^{\log_b y}=y\]
so we get here
\[\frac{xb^2}{a^2b^a}=b^{2-a}\]
Now we have
\[\frac{xb^{2-a}}{a^2}=b^{2-a}\]
\[\frac{x\cancel{b^{2-a}}}{a^2}=\cancel{b^{2-a}} 1\]
we get
\[x=a^2\]