What is the answer to 81^{5x-2}=27^{x+4} ?

The answer to 81^{5x-2}=27^{x+4} is x= 20/17

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81^{5x-2}=27^{x+4}

8 1^{5 x - 2} = 2 7^{x + 4}

x = \frac{20}{17}

Solution steps

8 1^{5 x - 2} = 2 7^{x + 4}

Apply exponent rules

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

Solve

4 (5 x - 2) = 3 (x + 4) : x = \frac{20}{17}

x = \frac{20}{17}

4^{x - 1} = 5

a, (x^{a})^{b} = b x^{a}

10000 = 40000 e^{- 0.06 t}

1 1^{x} = 1 1^{4} \cdot 1 1^{5}

lo g_{3} (7^{5 x}) = 10

What is the answer to 81^{5x-2}=27^{x+4} ?
The answer to 81^{5x-2}=27^{x+4} is x= 20/17
Solve for x: 81^{5x-2}=27^{x+4}
The solution is x= 20/17