Solved on Feb 10, 2024

Find the value of $\cosh(3\ln 3) - \sinh(3\ln 3)$ .

STEP 1

Assumptions
1. We are given the hyperbolic functions $\cosh(3x)$ and $\sinh(3x)$ .
2. We need to evaluate these functions at $x = \ln 3$ .
3. The hyperbolic cosine and hyperbolic sine functions are defined as: $\cosh x = \frac{e^x + e^{-x}}{2}$ $\sinh x = \frac{e^x - e^{-x}}{2}$
4. We will use properties of logarithms and exponentials to simplify the expressions.

STEP 2

First, we will write the expression $\cosh(3x) - \sinh(3x)$ using the definitions of $\cosh x$ and $\sinh x$ .
$\cosh(3x) - \sinh(3x) = \frac{e^{3x} + e^{-3x}}{2} - \frac{e^{3x} - e^{-3x}}{2}$

STEP 3

Now, we simplify the right-hand side by combining the fractions.
$\cosh(3x) - \sinh(3x) = \frac{e^{3x} + e^{-3x} - e^{3x} + e^{-3x}}{2}$

STEP 4

Next, we cancel out the $e^{3x}$ terms.
$\cosh(3x) - \sinh(3x) = \frac{2e^{-3x}}{2}$

STEP 5

Simplify the expression by dividing both the numerator and the denominator by 2.
$\cosh(3x) - \sinh(3x) = e^{-3x}$

STEP 6

Now we need to evaluate the expression at $x = \ln 3$ . Substitute $x$ with $\ln 3$ .
$\cosh(3\ln 3) - \sinh(3\ln 3) = e^{-3\ln 3}$

STEP 7

Recall the property of logarithms that states $e^{\ln a} = a$ . We will use this to simplify the exponent.
$e^{-3\ln 3} = (e^{\ln 3})^{-3}$

STEP 8

Apply the property of logarithms to simplify further.
$(e^{\ln 3})^{-3} = 3^{-3}$

STEP 9

Calculate the value of $3^{-3}$ .
$3^{-3} = \frac{1}{3^3}$

STEP 10

Calculate the value of $3^3$ .
$3^3 = 3 \times 3 \times 3 = 27$

STEP 11

Now, divide 1 by 27 to get the final value.
$3^{-3} = \frac{1}{27}$
The value of $\cosh(3x) - \sinh(3x)$ at $x = \ln 3$ is $\frac{1}{27}$ .

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Find the value of $\cosh(3\ln 3) - \sinh(3\ln 3)$ .

STEP 1

STEP 2

First, we will write the expression $\cosh(3x) - \sinh(3x)$ using the definitions of $\cosh x$ and $\sinh x$ .
$\cosh(3x) - \sinh(3x) = \frac{e^{3x} + e^{-3x}}{2} - \frac{e^{3x} - e^{-3x}}{2}$

STEP 3

Now, we simplify the right-hand side by combining the fractions.
$\cosh(3x) - \sinh(3x) = \frac{e^{3x} + e^{-3x} - e^{3x} + e^{-3x}}{2}$

STEP 4

Next, we cancel out the $e^{3x}$ terms.
$\cosh(3x) - \sinh(3x) = \frac{2e^{-3x}}{2}$

STEP 5

Simplify the expression by dividing both the numerator and the denominator by 2.
$\cosh(3x) - \sinh(3x) = e^{-3x}$

STEP 6

Now we need to evaluate the expression at $x = \ln 3$ . Substitute $x$ with $\ln 3$ .
$\cosh(3\ln 3) - \sinh(3\ln 3) = e^{-3\ln 3}$

STEP 7

Recall the property of logarithms that states $e^{\ln a} = a$ . We will use this to simplify the exponent.
$e^{-3\ln 3} = (e^{\ln 3})^{-3}$

STEP 8

Apply the property of logarithms to simplify further.
$(e^{\ln 3})^{-3} = 3^{-3}$

STEP 9

Calculate the value of $3^{-3}$ .
$3^{-3} = \frac{1}{3^3}$

STEP 10

Calculate the value of $3^3$ .
$3^3 = 3 \times 3 \times 3 = 27$

STEP 11

Now, divide 1 by 27 to get the final value.
$3^{-3} = \frac{1}{27}$
The value of $\cosh(3x) - \sinh(3x)$ at $x = \ln 3$ is $\frac{1}{27}$ .

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Find the value of cosh⁡(3ln⁡3)−sinh⁡(3ln⁡3)\cosh(3\ln 3) - \sinh(3\ln 3)cosh(3ln3)−sinh(3ln3).

STEP 1

STEP 2

STEP 3

Now, we simplify the right-hand side by combining the fractions.cosh⁡(3x)−sinh⁡(3x)=e3x+e−3x−e3x+e−3x2\cosh(3x) - \sinh(3x) = \frac{e^{3x} + e^{-3x} - e^{3x} + e^{-3x}}{2}cosh(3x)−sinh(3x)=2e3x+e−3x−e3x+e−3x​

STEP 4

Next, we cancel out the e3xe^{3x}e3x terms.cosh⁡(3x)−sinh⁡(3x)=2e−3x2\cosh(3x) - \sinh(3x) = \frac{2e^{-3x}}{2}cosh(3x)−sinh(3x)=22e−3x​

STEP 5

Simplify the expression by dividing both the numerator and the denominator by 2.cosh⁡(3x)−sinh⁡(3x)=e−3x\cosh(3x) - \sinh(3x) = e^{-3x}cosh(3x)−sinh(3x)=e−3x

STEP 6

Now we need to evaluate the expression at x=ln⁡3x = \ln 3x=ln3. Substitute xxx with ln⁡3\ln 3ln3.cosh⁡(3ln⁡3)−sinh⁡(3ln⁡3)=e−3ln⁡3\cosh(3\ln 3) - \sinh(3\ln 3) = e^{-3\ln 3}cosh(3ln3)−sinh(3ln3)=e−3ln3

STEP 7

Recall the property of logarithms that states eln⁡a=ae^{\ln a} = aelna=a. We will use this to simplify the exponent.e−3ln⁡3=(eln⁡3)−3e^{-3\ln 3} = (e^{\ln 3})^{-3}e−3ln3=(eln3)−3

STEP 8

Apply the property of logarithms to simplify further.(eln⁡3)−3=3−3(e^{\ln 3})^{-3} = 3^{-3}(eln3)−3=3−3

STEP 9

Calculate the value of 3−33^{-3}3−3.3−3=1333^{-3} = \frac{1}{3^3}3−3=331​

STEP 10

Calculate the value of 333^333.33=3×3×3=273^3 = 3 \times 3 \times 3 = 2733=3×3×3=27

STEP 11

Now, divide 1 by 27 to get the final value.3−3=1273^{-3} = \frac{1}{27}3−3=271​The value of cosh⁡(3x)−sinh⁡(3x)\cosh(3x) - \sinh(3x)cosh(3x)−sinh(3x) at x=ln⁡3x = \ln 3x=ln3 is 127\frac{1}{27}271​.

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Find the value of $\cosh(3\ln 3) - \sinh(3\ln 3)$ .

Now, we simplify the right-hand side by combining the fractions.
$\cosh(3x) - \sinh(3x) = \frac{e^{3x} + e^{-3x} - e^{3x} + e^{-3x}}{2}$

Next, we cancel out the $e^{3x}$ terms.
$\cosh(3x) - \sinh(3x) = \frac{2e^{-3x}}{2}$

Simplify the expression by dividing both the numerator and the denominator by 2.
$\cosh(3x) - \sinh(3x) = e^{-3x}$

Now we need to evaluate the expression at $x = \ln 3$ . Substitute $x$ with $\ln 3$ .
$\cosh(3\ln 3) - \sinh(3\ln 3) = e^{-3\ln 3}$

Recall the property of logarithms that states $e^{\ln a} = a$ . We will use this to simplify the exponent.
$e^{-3\ln 3} = (e^{\ln 3})^{-3}$

Apply the property of logarithms to simplify further.
$(e^{\ln 3})^{-3} = 3^{-3}$

Calculate the value of $3^{-3}$ .
$3^{-3} = \frac{1}{3^3}$

Calculate the value of $3^3$ .
$3^3 = 3 \times 3 \times 3 = 27$

Now, divide 1 by 27 to get the final value.
$3^{-3} = \frac{1}{27}$
The value of $\cosh(3x) - \sinh(3x)$ at $x = \ln 3$ is $\frac{1}{27}$ .