What is the answer to 27^{n^2}=9^{5n-4} ?

The answer to 27^{n^2}=9^{5n-4} is n=2,n= 4/3

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27^{n^2}=9^{5n-4}

2 7^{n^{2}} = 9^{5 n - 4}

n = 2, n = \frac{4}{3}

Decimal Notation

n = 2, n = 1.33333 \dots

Solution steps

2 7^{n^{2}} = 9^{5 n - 4}

Apply exponent rules

3 n^{2} = 2 (5 n - 4)

Solve

3 n^{2} = 2 (5 n - 4) : n = 2, n = \frac{4}{3}

n = 2, n = \frac{4}{3}

5^{x} = 62 5^{2}

7^{12 x} = 7^{2 x - 2}

8^{2 x} = \frac{1}{64}

25 6^{x} = 0.25

(\frac{1}{8})^{x} = 25

What is the answer to 27^{n^2}=9^{5n-4} ?
The answer to 27^{n^2}=9^{5n-4} is n=2,n= 4/3
Solve for n: 27^{n^2}=9^{5n-4}
The solution is n=2,n= 4/3