What is the answer to 128=(5^{2n-5})/(2^{4n)} ?

The answer to 128=(5^{2n-5})/(2^{4n)} is n=(7+5log_{2}(5))/(2(-2+log_{2)(5))}

Solve for n: 128=(5^{2n-5})/(2^{4n)}

The solution is n=(7+5log_{2}(5))/(2(-2+log_{2)(5))}

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128 = \frac{5 ^{2 n - 5}}{2 ^{4 n}}

n = \frac{7 + 5 lo g _{2} ( 5 )}{2 ( - 2 + lo g _{2} ( 5 ) )}

Solution steps

128 = \frac{5 ^{2 n - 5}}{2 ^{4 n}}

Apply exponent rules

7 = lo g_{2} (5) (2 n - 5) - 4 n

Solve

7 = lo g_{2} (5) (2 n - 5) - 4 n : n = \frac{7 + 5 lo g _{2} ( 5 )}{2 ( - 2 + lo g _{2} ( 5 ) )}

n = \frac{7 + 5 lo g _{2} ( 5 )}{2 ( - 2 + lo g _{2} ( 5 ) )}

Verify Solutions:

n = \frac{7 + 5 lo g _{2} ( 5 )}{2 ( - 2 + lo g _{2} ( 5 ) )}

True

The solution is

n = \frac{7 + 5 lo g _{2} ( 5 )}{2 ( - 2 + lo g _{2} ( 5 ) )}

2^{x} = 36

7^{3 x - 2} = 2401

5 (2^{x + 1}) - 4^{x} = 16

4 9^{x + 3} = 7^{x^{2} - 3 x - 8}

1 + 9^{x} = 3^{x + 1} + 3^{x - 1}

What is the answer to 128=(5^{2n-5})/(2^{4n)} ?
The answer to 128=(5^{2n-5})/(2^{4n)} is n=(7+5log_{2}(5))/(2(-2+log_{2)(5))}
Solve for n: 128=(5^{2n-5})/(2^{4n)}
The solution is n=(7+5log_{2}(5))/(2(-2+log_{2)(5))}