Question

If $4 + 6(e^{2x} + 1)\tanh x = 11\cosh x + 11\sinh x$, then $x =$

• A. $\log 10$
• B. $\log 4$
• C. $\log 5$
• D. $\log 2$

Hint:

Evaluate $\sinh x$, $\cosh x$ at $\log$ values using exponentials and verify identities numerically.

Answer 1

The Correct Option is D

Recall: $\tanh x = \dfrac{\sinh x}{\cosh x}$
Also, $e^{2x} + 1 = (\cosh x + \sinh x)^2$
So, LHS becomes: $4 + 6(e^{2x} + 1)\tanh x$
Try $x = \log 2$:
Then $e^x = 2 \Rightarrow e^{-x} = \dfrac{1}{2}$
$\cosh x = \dfrac{e^x + e^{-x}}{2} = \dfrac{2 + \frac{1}{2}}{2} = \dfrac{5}{4}$
$\sinh x = \dfrac{e^x - e^{-x}}{2} = \dfrac{2 - \frac{1}{2}}{2} = \dfrac{3}{4}$
So $\cosh x + \sinh x = \dfrac{5 + 3}{4} = 2$, square: $e^{2x} + 1 = 4 + 1 = 5$
$\tanh x = \dfrac{3/4}{5/4} = \dfrac{3}{5}$
Now check: LHS = $4 + 6 \cdot 5 \cdot \dfrac{3}{5} = 4 + 18 = 22$
RHS = $11(\cosh x + \sinh x) = 11 \cdot 2 = 22$
Hence, $x = \log 2$