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If \(\tanh (x)=\frac{1}{3}\), then \(\tanh (3 x)\) is
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Verified Answer
The correct answer is:
\(\frac{7}{9}\)
\(\begin{aligned}
& \tanh (x)=\frac{1}{3}, \tanh (3 x)=? \\
& \Rightarrow \quad \frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{1}{3} \Rightarrow 2 e^x=4 e^{-x} \Rightarrow\left(e^{2 x}=2\right) \\
& \text {and } \tanh (3 x)=\frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}}=\frac{\left(e^{2 x}\right)^3-1}{\left(e^{2 x}\right)^3+1}=\frac{8-1}{8+1}=\frac{7}{9}
\end{aligned}\)
& \tanh (x)=\frac{1}{3}, \tanh (3 x)=? \\
& \Rightarrow \quad \frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{1}{3} \Rightarrow 2 e^x=4 e^{-x} \Rightarrow\left(e^{2 x}=2\right) \\
& \text {and } \tanh (3 x)=\frac{e^{3 x}-e^{-3 x}}{e^{3 x}+e^{-3 x}}=\frac{\left(e^{2 x}\right)^3-1}{\left(e^{2 x}\right)^3+1}=\frac{8-1}{8+1}=\frac{7}{9}
\end{aligned}\)
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