What is the answer to solvefor m,2^m0.3^n0.72=1 ?

The answer to solvefor m,2^m0.3^n0.72=1 is m=(ln(\frac{25)/(18*0.3^n))}{ln(2)}

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solvefor m,2^m0.3^n0.72=1

solve for

m, 2^{m} 0. 3^{n} 0.72 = 1

m = \frac{ln ( \frac{25}{18 \cdot 0. 3 ^{n}} )}{ln ( 2 )}

Solution steps

2^{m} \cdot 0. 3^{n} \cdot 0.72 = 1

Multiply both sides by

100

72 \cdot 2^{m} \cdot 0. 3^{n} = 100

Divide both sides by

72 \cdot 0. 3^{n}

2^{m} = \frac{25}{18 \cdot 0. 3 ^{n}}

Apply exponent rules

m ln (2) = ln (\frac{25}{18 \cdot 0. 3 ^{n}})

Solve

m ln (2) = ln (\frac{25}{18 \cdot 0. 3 ^{n}}) : m = \frac{ln ( \frac{25}{18 \cdot 0. 3 ^{n}} )}{ln ( 2 )}

m = \frac{ln ( \frac{25}{18 \cdot 0. 3 ^{n}} )}{ln ( 2 )}

x, \frac{1}{a ^{x}} = e^{- y}

6 4^{2 x - 5} = 3 2^{x - 4}

n, 2 = t + t^{n}

\frac{( 2 ^{x} \cdot 3 ^{x} )}{( 1 2 ^{x} )} = \frac{1}{8}

32 = 8^{x}

What is the answer to solvefor m,2^m0.3^n0.72=1 ?
The answer to solvefor m,2^m0.3^n0.72=1 is m=(ln(\frac{25)/(18*0.3^n))}{ln(2)}