What is the answer to 3^{(x^2-4x-5)/(3x-1)}=1 ?

The answer to 3^{(x^2-4x-5)/(3x-1)}=1 is x=-1,x=5

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3^{(x^2-4x-5)/(3x-1)}=1

3^{\frac{x ^{2} - 4 x - 5}{3 x - 1}} = 1

x = - 1, x = 5

Solution steps

3^{\frac{x ^{2} - 4 x - 5}{3 x - 1}} = 1

Apply exponent rules

\frac{x ^{2} - 4 x - 5}{3 x - 1} ln (3) = ln (1)

Solve

\frac{x ^{2} - 4 x - 5}{3 x - 1} ln (3) = ln (1) : x = - 1, x = 5

x = - 1, x = 5

256 = (\frac{1}{4})^{3 x + 2}

t, x = e^{4 t}

2^{y} = \frac{1}{32}

170 = 180 (1 - e^{- 0.5 t})

1 0^{\frac{x ^{2} - 1}{x + 1}} = 10

What is the answer to 3^{(x^2-4x-5)/(3x-1)}=1 ?
The answer to 3^{(x^2-4x-5)/(3x-1)}=1 is x=-1,x=5
Solve for x: 3^{(x^2-4x-5)/(3x-1)}=1
The solution is x=-1,x=5