Solve for $x$ correct to two significant figures, the equation: $4^{2x+1}.5^{x-2}=6^{1-x}$ (Conflicting answer with book)
My method: $4^{2x}.4.5^{x}.5^{-2}=6.6^{-x} \Rightarrow \frac{4^{2x}.5^{x}}{25}=\frac{6.6^{-x}}{4} \Rightarrow \frac{4^{2x}.5^{x}}{6^{x}}=\log \left(37.5\right)$
$\Rightarrow \log \left(\frac{4^{2x}.5^{x}}{6^{x}}\right)=1.574$
$\Rightarrow \log \left(4^{2x}.5^{x}\right)-\log \left(6^{x}\right)=1.574 \Rightarrow 2x\log \left(4\right)+x\log \left(5\right)-x\log \left(6\right)=1.574$
$\Rightarrow 1.204x+0.699x-0.778x=1.574 \Rightarrow x = \frac{1.574}{1.125}=1.4$
Answer in book: $x=0.59$
Hint it should be $\log(16)$ rather than $4$. Then it'll be $x\log16+x\log5+x\log6=\log37.5$ so $2.66x=1.57$ giving $x=0.59$. Hope its clear.