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$\frac{2^x - 2^{-x}}{2^x + 2^{-x}} = \frac{1}{3}$ $2^x - 2^{-x} =1$ $2^x + 2^{-x} = 3$

My attempts

  1. A - b = 1

A + b = 3 Which lead to x = 1 and x = 0 inconsistency.

  1. $2^x - \frac{1}{2^x} = 1$

$2^x + \frac{1}{2^x} = 3$ Which lead to x = 1

Why two method gave different answers, why is it wrong?

Dini
  • 1,391

2 Answers2

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Actually for $$\dfrac ab=\dfrac cd$$

$\implies\dfrac ac=\dfrac bd=k$(say)

$k$ may not be $1$

Use https://www.qc.edu.hk/math/Junior%20Secondary/Componendo%20et%20Dividendo.htm

$$\dfrac{2^x}{2^{-x}}=\dfrac{3+1}{3-1}$$

$$2^{2x}=2^1\iff2^{2x-1}=1$$

Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$

lab bhattacharjee
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Both your methods are wrong since

$$\frac A B = \frac C D \not \Rightarrow A=C \quad \land \quad B=D$$

we have

$$\frac{2^x - 2^{-x}}{2^x + 2^{-x}} = \frac{1}{3} \iff \frac{4^x - 1}{4^x + 1} = \frac{1}{3}\iff 3\cdot4^x - 3=4^x + 1 \iff 2\cdot4^x=4 \implies x=\frac12$$

user
  • 154,566