What is the answer to sqrt(9^{2x+1)}=27^{x-5} ?

The answer to sqrt(9^{2x+1)}=27^{x-5} is x=16

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sqrt(9^{2x+1)}=27^{x-5}

9^{2 x + 1} = 2 7^{x - 5}

x = 16

Solution steps

9^{2 x + 1} = 2 7^{x - 5}

Apply exponent rules

2 x + 1 = 3 (x - 5)

Solve

2 x + 1 = 3 (x - 5) : x = 16

x = 16

2 5^{m - 2} = 5^{2 m + 4}

3^{2 x + 1} = 10 (3^{x}) - 3

(e^{4 x + 1})^{3} = e^{2}

65 = 1 5^{x}

62 5^{x + 5} = 5^{3 x - 4} 2 5^{2 x - 3}

What is the answer to sqrt(9^{2x+1)}=27^{x-5} ?
The answer to sqrt(9^{2x+1)}=27^{x-5} is x=16
Solve for x: sqrt(9^{2x+1)}=27^{x-5}
The solution is x=16