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Determine the values of 
x for which

64cosh^(4)x-64cosh^(2)x-9=0
Give your answers in the form 
q ln 2 where 
q is rational and in simplest form.

Determine the values of xx for which\newline64cosh4x64cosh2x9=064\cosh^{4}x-64\cosh^{2}x-9=0\newlineGive your answers in the form qln2q \ln 2 where qq is rational and in simplest form.

Full solution

Q. Determine the values of xx for which\newline64cosh4x64cosh2x9=064\cosh^{4}x-64\cosh^{2}x-9=0\newlineGive your answers in the form qln2q \ln 2 where qq is rational and in simplest form.
  1. Substitution Simplification: Let's start by simplifying the equation using a substitution. Let u=cosh2(x)u = \cosh^2(x). The equation then becomes 64u264u9=064u^2 - 64u - 9 = 0.
  2. Quadratic Equation Solution: Next, we'll solve the quadratic equation 64u264u9=064u^2 - 64u - 9 = 0. Using the quadratic formula, u=b±b24ac2au = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, where a=64a = 64, b=64b = -64, and c=9c = -9.
  3. Root Calculation: Plugging in the values, we get u=64±(64)2464(9)264u = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 64 \cdot (-9)}}{2 \cdot 64}. Simplifying inside the square root: u=64±4096+2304128u = \frac{64 \pm \sqrt{4096 + 2304}}{128}.
  4. Validating Solution: Further simplifying, u=(64±6400)/128=(64±80)/128u = (64 \pm \sqrt{6400}) / 128 = (64 \pm 80) / 128. This gives us u=144/128u = 144/128 or u=16/128u = -16/128, which simplifies to u=9/8u = 9/8 or u=1/8u = -1/8.
  5. Finding xx: Since cosh2(x)\cosh^2(x) cannot be negative, we discard u=18u = -\frac{1}{8}. We only consider u=98u = \frac{9}{8}. Now, we need to find xx such that cosh2(x)=98\cosh^2(x) = \frac{9}{8}.
  6. Square Root Calculation: Taking the square root on both sides, cosh(x)=98=38=324\cosh(x) = \sqrt{\frac{9}{8}} = \frac{3}{\sqrt{8}} = \frac{3\sqrt{2}}{4}.
  7. Equation Rearrangement: Using the definition of cosh(x)=ex+ex2\cosh(x) = \frac{e^x + e^{-x}}{2}, we set up the equation ex+ex2=324\frac{e^x + e^{-x}}{2} = \frac{3\sqrt{2}}{4}. Multiplying through by 22 and rearranging, we get ex+ex=322e^x + e^{-x} = \frac{3\sqrt{2}}{2}.
  8. Quadratic Equation Solution: Solving for exe^x, let y=exy = e^x. Then y+1y=322y + \frac{1}{y} = \frac{3\sqrt{2}}{2}. Multiplying through by yy, we get y2(322)y+1=0y^2 - \left(\frac{3\sqrt{2}}{2}\right)y + 1 = 0.
  9. Quadratic Equation Solution: Solving for exe^x, let y=exy = e^x. Then y+1y=322y + \frac{1}{y} = \frac{3\sqrt{2}}{2}. Multiplying through by yy, we get y2(322)y+1=0y^2 - \left(\frac{3\sqrt{2}}{2}\right)y + 1 = 0. Solving this quadratic equation for yy, we use the quadratic formula again: y=(322)±(322)242y = \frac{\left(\frac{3\sqrt{2}}{2}\right) \pm \sqrt{\left(\frac{3\sqrt{2}}{2}\right)^2 - 4}}{2}.

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