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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: $x=0.59$

AsukaMinato
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shredalert
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  • You get $2 \log(2) (2x + 1) + \log(5) (x-2) = \log(6)(1-x)$. After solving for $x$, I get the books answer. Used identities $\log(u v) = \log(u) + \log(v)$ and $\log(u^v) = v \log(u)$. – Moo Dec 07 '15 at 13:43
  • I used your method @idiot to get the answer. Could you please help me see where I went wrong in my method? – shredalert Dec 07 '15 at 13:50
  • I don't see how that is possible, since $6^{1-x}=6^1.6^{-x}$ – shredalert Dec 07 '15 at 13:54
  • Recall $\log(u^v) = v\log(u)$ – Moo Dec 07 '15 at 13:56

1 Answers1

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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.

Moo
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Archis Welankar
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