What is the answer to (3^x+3^{-x})/(3^x-3^{-x)}=20 ?

The answer to (3^x+3^{-x})/(3^x-3^{-x)}=20 is x=(ln(21/19))/(2ln(3))

Solve for x: (3^x+3^{-x})/(3^x-3^{-x)}=20

The solution is x=(ln(21/19))/(2ln(3))

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\frac{3 ^{x} + 3 ^{- x}}{3 ^{x} - 3 ^{- x}} = 20

x = \frac{ln ( \frac{21}{19} )}{2 ln ( 3 )}

Decimal

x = 0.04554 \dots

Solution steps

\frac{3 ^{x} + 3 ^{- x}}{3 ^{x} - 3 ^{- x}} = 20

Multiply both sides by

3^{x} - 3^{- x}

\frac{3 ^{x} + 3 ^{- x}}{3 ^{x} - 3 ^{- x}} (3^{x} - 3^{- x}) = 20 (3^{x} - 3^{- x})

Simplify

3^{x} + 3^{- x} = 20 (3^{x} - 3^{- x})

Apply exponent rules

3^{x} + (3^{x})^{- 1} = 20 (3^{x} - (3^{x})^{- 1})

Rewrite the equation with

3^{x} = u

u + (u)^{- 1} = 20 (u - (u)^{- 1})

Solve

u + u^{- 1} = 20 (u - u^{- 1}) : u = \frac{21}{19}, u = - \frac{21}{19}

u = \frac{21}{19}, u = - \frac{21}{19}

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = \frac{21}{19} : x = \frac{ln ( \frac{21}{19} )}{2 ln ( 3 )}

Solve

3^{x} = - \frac{21}{19} :

No Solution for

x \in R

x = \frac{ln ( \frac{21}{19} )}{2 ln ( 3 )}

Verify Solutions:

x = \frac{ln ( \frac{21}{19} )}{2 ln ( 3 )}

True

The solution is

x = \frac{ln ( \frac{21}{19} )}{2 ln ( 3 )}

4 \times e^{2.7 x} = 33

4^{x + 2} = 8

e^{2 x} - 8 e^{x} = 20

y, ln (x) + e^{- \frac{y}{x}} = c

3^{x + 1} + 3^{x - 1} = 30

What is the answer to (3^x+3^{-x})/(3^x-3^{-x)}=20 ?
The answer to (3^x+3^{-x})/(3^x-3^{-x)}=20 is x=(ln(21/19))/(2ln(3))
Solve for x: (3^x+3^{-x})/(3^x-3^{-x)}=20
The solution is x=(ln(21/19))/(2ln(3))