What is the answer to 1/(81^{(x-2))}=27^{(1-x)} ?

The answer to 1/(81^{(x-2))}=27^{(1-x)} is x=5

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1/(81^{(x-2))}=27^{(1-x)}

\frac{1}{8 1 ^{(x - 2)}} = 2 7^{(1 - x)}

x = 5

Solution steps

\frac{1}{8 1 ^{(x - 2)}} = 2 7^{(1 - x)}

Apply exponent rules

4 (- x + 2) = 3 (1 - x)

Solve

4 (- x + 2) = 3 (1 - x) : x = 5

x = 5

Verify Solutions:

x = 5

True

The solution is

x = 5

e^{12 x - 1} = (e^{10})^{x}

3^{x} + x = 11

3^{2 - x} = 3^{x^{2}}

8^{x + 1} + 2^{3 x - 1} = \frac{17}{16}

9^{2 x} - 3 \cdot 9^{x} + 2 = 0

What is the answer to 1/(81^{(x-2))}=27^{(1-x)} ?
The answer to 1/(81^{(x-2))}=27^{(1-x)} is x=5
Solve for x: 1/(81^{(x-2))}=27^{(1-x)}
The solution is x=5